Sir, I have another solution of this problem but I don't know whether it is correct or not. However it is as follows Given that a+b+c=0. and we have to find the value of ( b-c/a +c-a/b +a-b/c)(a/b-c +b/c-a +c/a-b) Let b-c/a =x c-a/b=y a-b/c=z From above we get b-c=ax......(1) c-a=by.....(2) a-b=cz......(3) Adding 1,2&3 we get ax+by+cz=0........(4) It's already given that a+b+c=0.....(5) From eq.4&5 we can assume that x=y=z=1 Now replacing b-c/a, c-a/b & a-b/c by x, y & z we get the expression in the question as (x+y+z)( 1/x + 1/y + 1/z) =(1+1+1)(1/1+1/1+1/1) as x,y,z is 1 =3×3 =9
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
Sir I got the answer by multiplying both the brackets then simplifying two terms of them and six were identical terms so just wrote down same for other 4 and the remaining term was +3 and I got 9 by simplifying all the terms. Maza aa gaya sir... 😀
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
Hello sir Maine is question attempt diya khud se aur bahut hi aasani se ho gaya.. Isame a+b+c=0 hai to a+b=-c ho jayega. Aur is chiz pehale me c ke jagah par put Kar denge.. To sab kat ke -1 aa jayega aur usi Tarah se sab -1 aa jayega kul mila ke (-3)×(-3) =9
It was an amazing question sir. Really it gives so much satisfaction when we get answer of these algebraic questions. Please keep uploading more of such videos😊
Sir honestly speaking I have attempted this question by 4 ways which I have in my mind but didnot get the answer they become very complex in the end .... But randomly I put 😂 a=3/4 b=-1/2 c=-1/4 And solve it within a second ❤❤❤❤❤❤❤❤🎉🎉🎉🎉🎉
there are 2 brackets in first bracket, b=c then a=-2b=-2c putting these values first brackett becomes zero that means a=b,b=c,c=a are factors of first brackett and since its homogeneous fraction throughout so merely looking at it one can tell, that finally first brackett is of the form k (a-b)(b-c)(c-a)/abc now look at second brackett a=0 means b=-c which makes second brackett =0 hence second brackett which has all terms homogeneous will be m abc/(a-b)(b-c)(c-a) multiplying both brackets all the terms having a,b,c will cancel out. so till here merely looking at problem with logic you know it is constant. means invariant, now put a=2, b=1, c=-1 since we have already proved it is invariant or constant, which makes it 9 so you don’t need pen paper for this
homogeneity and cyclicity makes such problems very easy, as discussed in detailed solution I have written above. just use homogeneity cyclicity and invariance
You can't put a = 0 as it's not in the domain right ? In the first bracket a is in the denominator . Except for zero we can use any other value and your explanation is nice then .
Sir I have a doubt in a question please tell 🙏🙏🙏🙏 Please sir Question) prove that 1 - cos²A - cos²B - cos²C -2cosA.cosB.cosC = 4sin( A+B+C/2 ).sin (A+B-C/2).sin ( A-B+C/2).sin(C+B-A/2) Please tell sir
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
I solved it orally by inspection and answer is 9. Tell me if something is wrong with the reasoning. Given condition as well as expression are symmetric w.r.t a,b,c. So there is no reason that c-a/b could be different value than b-c/a etc. Hence, first factor in expression should be written as 3 x (b-c)/a. Similarly second becomes, 3 x a/(b-c). Product is 9.
b-c/a=x c-a/b =y a-b/c=z Make this type of (x+y+z)1/x+1/y+1/z) eq 1 b-c=ax c-a=by a-b =zy add up all these then ax+by+cz=0 then ax +by +cy =a+b+c a(x-1)+b(y-1)+c(z-1)=0 Therefore x= y=z=1 Then put the value in eq 1 3*3=9 Is it right way to do it ?? If not please correct me
Sabhi me plus one karege aur plus 3 means minus bhi karege both equation and finally solve karege to numerator me a+b+c milega aur uski value zero hi finally minus 3 and minus 3 milega to plus 9 hoga
Dear Sir mere pass two questions hai Pythagoras se related ap ke liye first ki world ka sabase chota area wala right triangle kon sa hai jiska area sabase Chhota ek natural number ho aur Pythagoras (a²+b²=c²) ko satisfy bhi kare note a,b,c irrational number na ho.for example 2²+1²=(√5)² yaha par to area 1 jo ki ek sabase chhota natural number toh hai lekin 'c' irrational number hai tino rational number yani tino p/q ke form me ham likh paye hona chahiye.second question 3²+4²=5² area 6 but see that( 7/10)²+(120/7)²=(1201/70)² isaka bhi area 6 hi hai kya ap ek aur(a,b,c) rational length bata sakate hai jiska area 6 hi ho aur a²+b²=c² ho jaha a,b,c, rational ho aur 1/2×ab=6 ho.
Let take , a=b-c ,b=c-a & c=a-b Then it satisfied a+b+c=0 Then put the value of only a , only b & only c the the fraction will cancel out and comes as 3*3 Which results 9
But, how you have taken a=b-c , b=c-a , c=a-b, Since, it is given that a+b+c=0, so, a = -b-c ; b= -c-a ; c = -a-b , so you cannot take the values as stated above. And if you say that after adding all the 3 equations, you are getting 0, then you should observe that *b-c+c-a+a-b=0* and not *a+b+c=0*, and you should observe that both of these are different equations. So, this can't be the observation for this question.
Sir I have a doubt in a question please tell 🙏🙏🙏🙏 Please sir Question) prove that 1 - cos²A - cos²B - cos²C = 4sin( A+B+C/2 ).sin (A+B-C/2).sin ( A-B+C/2).sin(C+B-A/2) Please tell sir
best way to solve this problem without putting the values was to multiply the terms simply and then add the terms with same denominator which could easily be solved by seeing the symmetry by this approach we can get rid of boring factorization
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
@@ashugarg3513coincidence. A+B+C = 0 means they are negative and positive. (As all 3 can't be 0) and AM GM HM is only applicable for positive integer I tried with that and yes u get the answer 9 (when equality holds) it only holds when b-c / a = c-a /b = a-b/c
Sir i have an easy solution Since a+b+c=0 so b-c/a= b-c/-b-c=-(b+c/b+c)=-1 c-a/b=c-a/-c-a=-(c+b/c+b)=-1 a-b/c=a-b/-a-b=-(a+b/a+b)=-1 And a/b-c=-b-c/b-c=-(b+c/b+c)=-1 b/c-a=-c-a/c-a=-(c+a/c+a)=-1 c/a-b=-a-b/a-b=-(a+b/a+b)=-1 Now putting this value in equation [-1+(-1)+(-1)]×[-1+(-1)+(-1)] [-1-1-1]×[-1-1-1] [-3]×[-3]=9
What are you saying??? b-c/a = b-c/-b-c = -(b+c/b+c)??? Do check your dumb solution before posting it. Try multiplying (-1) to (b+c) it will result in -b-c not b-c.
First I take the values of a,b,c from equation ( a+b+c=0),and then putting into the ques,then I got answer 9 easily in first attempt in the begining of the video
Sir, If we take let, a=k b=2k c=-3k then , a+b+c=k+2k+(-3k) =0 Now,if we use these value In following question then we get ,9 ans My question is that This method may be consider as an another method, Please reply me if possible 🙏 This is subjective method or not
The approach may give you the answer but imagine if the addition was in form of roots or complex numbers then assuming the variable K won't work, so if Mcq type comes you can apply your trick but in integer type you have to go with this method
Sir i have got the same answer by putting the values of a,b,c from given equation.....Than by simplifying each fraction by componendo and dividendo.....after that I have got a perfect square in the form -(c/b + a/c + b/a)^2 by taking lcm i have got a quadratic equation in the numerator....and if a+b+c =0 then there is a possibility that a=b=c then by putting these value in numerator.....i have got (a^3+b^3+c^3 /abc)^2 then i have got answer as 9 Jee 2025 aspirant ❤️
Just calculate the value for each using expression a+b+c = 0 a = -b -c b = -a -c c = -a -b Put this all values in given question and you'll get 9 The answer
SIR!!!! I solved this question in first attempt 😀😀🤯🤯 I just applied the general approach by taking LCM and so on and I unbelievable solved the question in 1st attempt. Sir you are the greatest mathematics teacher of all time And my inspiration is Ramanujan sir ❤❤ I am in class 10th Thank you sir
@Manoj-sd3yi Thanks I appreciate it But you know I am not kind of intrested in Olympiad. I am interested in JEE ADVANCED and ngl I was able to solve some of JEE ADV. Trigonometry question (I am not lying) Also I usually don't see people like you who encourage the young bloods instead of making fun of them or becoming jealous of them. My best wishes to you sir have a great day Thanks
@@a.kverma_6625 u are really 10 class kid bro its sad that some people get to hear name of jee in class 12 and kids like u starts solving in class 10 😥😥
Ek Arsa beet Gaya lekin Aman sir Aaj bhi best hi hai❤❤❤😊😊😊😊😊😊
Sir, I have another solution of this problem but I don't know whether it is correct or not. However it is as follows
Given that a+b+c=0. and we have to find the value of
( b-c/a +c-a/b +a-b/c)(a/b-c +b/c-a +c/a-b)
Let b-c/a =x
c-a/b=y
a-b/c=z
From above we get
b-c=ax......(1)
c-a=by.....(2)
a-b=cz......(3)
Adding 1,2&3 we get
ax+by+cz=0........(4)
It's already given that a+b+c=0.....(5)
From eq.4&5 we can assume that
x=y=z=1
Now replacing b-c/a, c-a/b & a-b/c
by x, y & z we get the expression in the question as
(x+y+z)( 1/x + 1/y + 1/z)
=(1+1+1)(1/1+1/1+1/1) as x,y,z is 1
=3×3
=9
nice one bro
I think it's wrong.. its correct to prove that x=y=z however you can't say they are equal to one. They can be any number.
@@kindafool4083yeah but u can do 3x*3/x=9 n the method works
@@Therapoetic right, yeah
Wrong imagine if x was 1 it means b-a=c
sir i solved it in first attempt
and it was a nice experience
keep bringing such question
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
@@zeroplays9915it is mentioned in Sl loney
Select the column matrics takin column matrics 1,2,-3 or 2,3 -5 al determinant becomes 9 as substitute and two multiples.
My most favourite maths teacher on UA-cam. Lots and lots of thanks sir ❤❤😊
The satisfaction you well when you solve the question in your first try.. Is unmatchable❤.. 🙃
Sir I got the answer by multiplying both the brackets then simplifying two terms of them and six were identical terms so just wrote down same for other 4 and the remaining term was +3 and I got 9 by simplifying all the terms.
Maza aa gaya sir... 😀
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
@@zeroplays9915 tum arjuna 1.0 se ho na?
@@zeroplays9915 good approach brother
@@reenabenpatel7916 no bro mai VMC ka hu
I did it by using Am>= Hm inequality .. one step answer 😂
Short trick
(No. Of variable)^2
3^2
9
🙀
😂
I assumed a=1, b=2 and c=-3. The simplified the expression by putting their values and got 9. 😊😊
That's pure algebra
That's the wrong way of doing it
Try again bro
@@a.kverma_6625 Ya because this question was subjective so this method will be wrong..
This method can be used when it is objective, but in subjective this method will be invalid
The mighty assumtion method
Simply put -a-b in place of -c (from above equation), similarly do for -a and -b . We will get the answer in 2 steps
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
@@zeroplays9915han bhai thik he samj Gaye 10 bar ek cheez ko sabke comment pe kyun likh rahe ho
Substitute a=3, b=-2, c=-1 or anything (not all zero) which makes a+b+c=0 such that no two of them are same.
ssc?
U can also put the value of b-c,c-a from the given equation a+b+c=0 from where by simplifying we get 9 answer
keep doing what you are doing...really best on you tube.. you literally teach straight from the heart...
Sir yeh question amu mai pochaa jata hai entrance test msi
Hello sir Maine is question attempt diya khud se aur bahut hi aasani se ho gaya..
Isame a+b+c=0 hai to
a+b=-c ho jayega.
Aur is chiz pehale me c ke jagah par put Kar denge..
To sab kat ke -1 aa jayega aur usi Tarah se sab -1 aa jayega kul mila ke (-3)×(-3)
=9
Let
a= -1 , b= -2 , c=3
2 minutes method
BEST example of essay writing..try to see substance of question..think first
Take values of a b and c as -1 -1 and 2
Nhi kar sakte bro
U can't take as there is a-b in denominator so it will become something upon zero
Legend of Mathematical world 🌍
Simple sir 👍❤️😍
SIR YOU ARE GENIUS OF MATHS
Ye question Allen ki Class 9 Advance Surds and Indices ke 2016 version me diya h.. Uske baad se toh hata diya..
One of the most tricky question i've seen so far. Got it on the second try btw.
We would love to see more questions like these.
Love you sir❤❤❤
It was an amazing question sir.
Really it gives so much satisfaction when we get answer of these algebraic questions. Please keep uploading more of such videos😊
Sir honestly speaking I have attempted this question by 4 ways which I have in my mind but didnot get the answer they become very complex in the end ....
But randomly I put 😂 a=3/4
b=-1/2
c=-1/4
And solve it within a second ❤❤❤❤❤❤❤❤🎉🎉🎉🎉🎉
😂
Question to bahut easy tha😂😂
Sir I am very happy because I solve this by first attempt
there are 2 brackets
in first bracket, b=c then a=-2b=-2c
putting these values
first brackett becomes zero
that means
a=b,b=c,c=a are factors of first brackett
and since its homogeneous fraction throughout so merely looking at it one can tell, that finally first brackett is of the form
k (a-b)(b-c)(c-a)/abc
now look at second brackett a=0 means b=-c
which makes second brackett =0
hence second brackett which has all terms homogeneous will be
m abc/(a-b)(b-c)(c-a)
multiplying both brackets all the terms having a,b,c will cancel out.
so till here merely looking at problem with logic you know it is constant.
means invariant, now put a=2, b=1, c=-1 since we have already proved it is invariant or constant, which makes it 9
so you don’t need pen paper for this
homogeneity and cyclicity makes such problems very easy, as discussed in detailed solution I have written above.
just use homogeneity cyclicity and invariance
You can't put a = 0 as it's not in the domain right ? In the first bracket a is in the denominator . Except for zero we can use any other value and your explanation is nice then .
@@genos1458 yes, my bad, put any number other than that, Thankyou
@@genos1458 edited
Sir ur the best teacher of maths
Sir I have a doubt in a question please tell 🙏🙏🙏🙏
Please sir
Question) prove that
1 - cos²A - cos²B - cos²C -2cosA.cosB.cosC = 4sin( A+B+C/2 ).sin (A+B-C/2).sin ( A-B+C/2).sin(C+B-A/2)
Please tell sir
put a=1/3, b=2/3, c=-1
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
Could you elaborate a bit
There is another method.... kind of silly one but works!!....
Just put a=1 b=2 c=(-3)....
Ans is 9....
I did it within a minute....😬😬
What a beautiful question sir bring more such tough questions solutions
I solved it orally by inspection and answer is 9. Tell me if something is wrong with the reasoning. Given condition as well as expression are symmetric w.r.t a,b,c. So there is no reason that c-a/b could be different value than b-c/a etc. Hence, first factor in expression should be written as 3 x (b-c)/a. Similarly second becomes, 3 x a/(b-c). Product is 9.
b-c/a=x c-a/b =y a-b/c=z
Make this type of (x+y+z)1/x+1/y+1/z) eq 1
b-c=ax c-a=by a-b =zy add up all these then
ax+by+cz=0 then
ax +by +cy =a+b+c
a(x-1)+b(y-1)+c(z-1)=0
Therefore x= y=z=1
Then put the value in eq 1
3*3=9
Is it right way to do it ?? If not please correct me
How it became x=y=z= 1
a=3,b=-2,c=-1 b-c not equal to c-a not equal to a-b not equal to 0
hence upon substitution we get
9
I am going in motion coaching i am in class 9th and this question is in our 2nd exericse
In each bracket we can operate with +3 and -3 we get -3 in both brackets which on multiplication get 9 💯
I also got the same.
kha se laate ho itna dimag
kuch bhi karna nhi h isme toh.
such a beautiful subject is maths
you can make solution smaller to very longer as you want.
@@himanshubaliyan5015Question is subjective
Sabhi me plus one karege aur plus 3 means minus bhi karege both equation and finally solve karege to numerator me a+b+c milega aur uski value zero hi finally minus 3 and minus 3 milega to plus 9 hoga
Dear Sir mere pass two questions hai Pythagoras se related ap ke liye first ki world ka sabase chota area wala right triangle kon sa hai jiska area sabase Chhota ek natural number ho aur Pythagoras (a²+b²=c²) ko satisfy bhi kare note a,b,c irrational number na ho.for example 2²+1²=(√5)² yaha par to area 1 jo ki ek sabase chhota natural number toh hai lekin 'c' irrational number hai tino rational number yani tino p/q ke form me ham likh paye hona chahiye.second question 3²+4²=5² area 6 but see that( 7/10)²+(120/7)²=(1201/70)² isaka bhi area 6 hi hai kya ap ek aur(a,b,c) rational length bata sakate hai jiska area 6 hi ho aur a²+b²=c² ho jaha a,b,c, rational ho aur 1/2×ab=6 ho.
Let take ,
a=b-c ,b=c-a & c=a-b
Then it satisfied a+b+c=0
Then put the value of only a , only b & only c the the fraction will cancel out and comes as 3*3
Which results 9
But, how you have taken
a=b-c , b=c-a , c=a-b,
Since, it is given that a+b+c=0, so, a = -b-c ; b= -c-a ; c = -a-b , so you cannot take the values as stated above.
And if you say that after adding all the 3 equations, you are getting 0, then you should observe that *b-c+c-a+a-b=0* and not *a+b+c=0*, and you should observe that both of these are different equations. So, this can't be the observation for this question.
Sir I have a doubt in a question please tell 🙏🙏🙏🙏
Please sir
Question) prove that
1 - cos²A - cos²B - cos²C = 4sin( A+B+C/2 ).sin (A+B-C/2).sin ( A-B+C/2).sin(C+B-A/2)
Please tell sir
Sir please teach for olympiads
You are the best maths teacher
Hello sir,, kabhi NDA ke important questions par aesi video aane chahiye..
Maza aagya sir ❤❤ mujse solve hogya
Sir plzzzz 11th ke sare chapters ke lecture dobara phir se naye Wale bana do plzzzzz sirrr...
Agar chahiye to paid batch kar lijiye
Sir aap ka phone number kya he
2:36 khud ban gya maja bhi aagya
Mera tarika alag tha par vo bhi sahi tha
First sir love you bhannaat❤❤❤k
best way to solve this problem without putting the values was to multiply the terms simply and then add the terms with same denominator which could easily be solved by seeing the symmetry by this approach we can get rid of boring factorization
i did it using trignometry, a=tan(x), b=tan(y), c=tan(z) and assuming x+y+z=180deg, so [tan(x)][tan(y)][tan(z)] = tan(x)tan(y)tan(z) and tan(x/2)tan(y/2) + tan(y/2)tan(z/2) + tan(z/2)tan(x/2) = 1 and just put in a b c in the equation in form of tan and got the answer.
No best way is to use AM >= HM inequality
@@ashugarg3513coincidence. A+B+C = 0 means they are negative and positive. (As all 3 can't be 0) and AM GM HM is only applicable for positive integer I tried with that and yes u get the answer 9 (when equality holds) it only holds when b-c / a = c-a /b = a-b/c
Humare coaching ke teacher ne notes mein likhwaaya!! JEE!!
Thanks a lot sir 🙏 ❤
Question karka maza aa gaya sir..... 👍👍❤❤😇😇
Maine ye question 2 minute me bna liya 1 page only 😅😅😅 real
We can just assume a=3, b=-2, c=-1 then solve it and easily it will solve
Sir maja aa gya please aise questions aur layega
1st attempt me ho gya
Olympiad batch start kro n sir❤ je
Sir consider a+b+c=1+w+w^2=0
And put value in equation you will get 9 answer
Sir i have an easy solution
Since a+b+c=0
so
b-c/a= b-c/-b-c=-(b+c/b+c)=-1
c-a/b=c-a/-c-a=-(c+b/c+b)=-1
a-b/c=a-b/-a-b=-(a+b/a+b)=-1
And
a/b-c=-b-c/b-c=-(b+c/b+c)=-1
b/c-a=-c-a/c-a=-(c+a/c+a)=-1
c/a-b=-a-b/a-b=-(a+b/a+b)=-1
Now putting this value in equation
[-1+(-1)+(-1)]×[-1+(-1)+(-1)]
[-1-1-1]×[-1-1-1]
[-3]×[-3]=9
What are you saying???
b-c/a = b-c/-b-c = -(b+c/b+c)???
Do check your dumb solution before posting it.
Try multiplying (-1) to (b+c) it will result in -b-c not b-c.
@@diteshsingh9588 It's my fault I didn't check 😅
put a=3,b=-2 and c=-1 and find value
Yyeee.......I have solved this question and I feel very happy 😊🎉
Sir,maja hi a gya........
Sir you are really greate😁😁😁😁🎉🎉🎉🎉🎉
Sir simply AM>= HM kyu nahi laga sakte usse bhi answer 9 hi ara hai
Saare number positive hone chie jo yha nai h
First I take the values of a,b,c from equation ( a+b+c=0),and then putting into the ques,then I got answer 9 easily in first attempt in the begining of the video
Sir,
If we take let, a=k
b=2k
c=-3k
then ,
a+b+c=k+2k+(-3k)
=0
Now,if we use these value
In following question
then we get ,9 ans
My question is that
This method may be consider
as an another method,
Please reply me if possible 🙏
This is subjective method or not
The approach may give you the answer but imagine if the addition was in form of roots or complex numbers then assuming the variable K won't work, so if Mcq type comes you can apply your trick but in integer type you have to go with this method
Sabse aasan a=3 b=-2 c = -1 put karo ans aa jayega
Simply put 1 2 and -3
bro a-b = c aise lekar sab ko divite karke 2 side me 1+1+ 1 karle 3*3 = 9 ho jaiga itni mehenot karni ki keya jorurot he😁
Sir ek international Olympiad mai ek windmill wala ques krwa do apke style may maja aa jayega
Just put 3 , -2,-1
And calculate answer
A=1b=2 c=-3 finish
What a brilliant solutions of a typical expressions
Many ways to solve this problem. This method is classic algebra love it..
Sir i have got the same answer by putting the values of a,b,c from given equation.....Than by simplifying each fraction by componendo and dividendo.....after that I have got a perfect square in the form -(c/b + a/c + b/a)^2 by taking lcm i have got a quadratic equation in the numerator....and if a+b+c =0 then there is a possibility that a=b=c then by putting these value in numerator.....i have got (a^3+b^3+c^3 /abc)^2 then i have got answer as 9
Jee 2025 aspirant ❤️
It's wrong if a+b+c=0,then possibility can be a=b=c=0, not a=b=c
@@Rishabh-wc6fv both are the possibilities
Bro i got it in first attempt
Class 9th
I just did am-hm and predicted it would be 9 in just 3 Seconds as these type of questions usually have answer which is minimum or maximum
put w and w^2 and 1 you will get the answer
Just calculate the value for each using expression
a+b+c = 0
a = -b -c
b = -a -c
c = -a -b
Put this all values in given question and you'll get 9
The answer
A problem with many solutions
I also used this
Easy question tha.... 9 hai answer ....btw thanks for boosting my confidence... simply brackets ko multiply bhi kar sakte hai
Yeh question to class 10 me tha hamara😂
Let a=1 and b=2 and c=-3 taking put the question and solv it then answer is 9
Maza aagaya.... Got the answer!
Feels happy in being in that 1% people
Just put any 3 number satisfying A+B+C=0 and solve it
Example-
A=1
B=2
C=-3
Sir kya hum ise a b aur c ki koi value rakh kar solve kar sakte
Sir maja aa gaya factor karne ke baad halva ho gaya
Sir maine bina open kiye cyclic se factor kiya
Put a =1 , b =2 and c = -3
Sir isme agar(a-b/c + b-c/a + c-a/b mein har ek mein +1 karde toh sirf do steps mein answer aa jayega
sir maine first attempt mei 6 nikal diya with expanding and got 6 due to great mistake
sir please do gaokao and integral bee series
Kal mera chemistry ka test hai or ye maths ka addiction chuth hi nahi raha
Sir kya ham iska answer random number likhke kar sakte hai kya exam me?
Sir it can be done very easily by a3+b3+c3 = 3abc identity, because a+b+c = 0
So according to you a+b=O And and Accordingly to you A^2+b^2=O ?
@@Manaschoudhary3636 nahh, it is an identity I'm not saying a+b = 0 ,it is given in the question that a+b+c=0 so we can use the identity I mentioned
I am in class 9 and i solve this in my 2nd attempt love ur videos sir ❤️❤️
Bohot easy tha but maine thoda lengthy hogaya aur ans 9 aya but bohot hi maja ayaaaa sirji😻😻😻🎉🎊🙅
Thank you sir very much
SIR!!!!
I solved this question in first attempt 😀😀🤯🤯
I just applied the general approach by taking LCM and so on and I unbelievable solved the question in 1st attempt.
Sir you are the greatest mathematics teacher of all time
And my inspiration is Ramanujan sir ❤❤
I am in class 10th
Thank you sir
Juta kaha hai mera
@@toxicgaming2020Apna pichwada check kar mil jayega.
Aur agar barf wagera chahiye hoga to bol dena kyuki teri bahut jyada hi jal rahi hai
@@a.kverma_6625 oh my god bhai bura maan gya
@Manoj-sd3yi Thanks I appreciate it
But you know I am not kind of intrested in Olympiad. I am interested in JEE ADVANCED and ngl I was able to solve some of JEE ADV. Trigonometry question (I am not lying)
Also I usually don't see people like you who encourage the young bloods instead of making fun of them or becoming jealous of them.
My best wishes to you sir have a great day
Thanks
@@a.kverma_6625 u are really 10 class kid bro its sad that some people get to hear name of jee in class 12 and kids like u starts solving in class 10 😥😥
Sir maza to apko dekhe ke hi aa jata hai
I have solved it
Ans 9
Bohat maja aya
My friends trying their best to solve this by all type of calculation
Le me putting
a=1
b=2
c=-3 🎉🎉🎉🎉