A bit easier (with the same idea) is to just complete the cube on both equations: with u = a-1, v = b-1 we arrive at u³+2u-14 = 0, v³+2v+14 = 0, adding the two and factoring yields (u+v)(u²+uv+v²+2) = 0 and with the quadratic having no real roots we arrive at u+v = a+b-2 = 0 hence a+b = 2
Rearranging (i.e. completing the cube) the two equations become : (a-1)^3 +2(a-1)-14=0 and (b-1)^3 + 2(b-1) +14 =0. If a-1 is solution to the first then -(a-1) is a solution to the second. So b-1= -(a-1) that is a+b =2.
Adding them and completing the cube gave me: (a+b)^3 - 3(a+b)^2 + 5(a+b) - 6 = 3ab(a+b) - 6ab if a + b = u then: u^3 - 3u^2 + 5u - 6 = 3ab(u - 2) which led me to think: "if u = 2 then i get rid of the ab on the RHS, so let's test if u=2 is a solution" in other words, I checked that value because it would be convenient for me to have a solution there and when I checked it, I found it's a solution not the most satisfying method, but it is proof that a+b = 2 is a valid solution
my method: i added the two equations to get a³+b³-3(a²+b²)+5(a+b)-6=0 now i rewrites all terms with a+b and ab and got (a+b)((a+b)²-3ab)-3((a+b)²-2ab)+5(a+b)-6=0 and substituted a+b=u and ab=v and got finally u³-3u²+5u-3uv+6v-6=0 which easily factors in (u-2)(u²-u-3v+3)=0 from where u=2 but I is nothing but a+b=2 and the problem is that I don't know what to do with the second bracket.
Substitute u= a + b and v = ab in the second bracket. Then, (a+b)^2 - ab - 3ab + 3 = 0 => (a-b)^2 + 3 = 0 => (a-b)^2 = -3 Which does not give real solution for all a,b belongs to R. So, we can simply ignore this result. This is what I think we should do with the other bracket
If you complete the cube in the 1st and 2nd eqn you will get (a-1)^3 + 2(a-8)=0 and for the 2nd (b-1)^3 + 2(b+6)=0 if you add them you get (a-1)^3+(b-1)^3 +2(a-8) +2(b+6)=0, factor the sum of cubes and pull out a two and you get (a+b-2)(a^2 +b^2 -ab -a -b+3)=0 from this you get a+b=2 or a^2 +b^2 -ab -a -b +3=0 multiplying by two and completing the squares gives you, (a-b)^2 +(a-1)^2+(b-1)^2 +4=0, therefore no more real sol. for a,b.
@@piyush-kx6vi I am sorry I can’t use telegram,try to know the numerical ans I suppose the question lacks some info and is based on transformation of roots.
A bit easier (with the same idea) is to just complete the cube on both equations: with u = a-1, v = b-1 we arrive at u³+2u-14 = 0, v³+2v+14 = 0, adding the two and factoring yields (u+v)(u²+uv+v²+2) = 0 and with the quadratic having no real roots we arrive at u+v = a+b-2 = 0 hence a+b = 2
v=b-1 I think?
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
@@dneary Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
@@piyush-kx6vi I'd need to think about it, but the factorization of x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx) looks promising.
@@dnearyhere a^1/3 is give. I think it is hard to solve. Plz help me to solve this
Rearranging (i.e. completing the cube) the two equations become : (a-1)^3 +2(a-1)-14=0 and
(b-1)^3 + 2(b-1) +14 =0. If a-1 is solution to the first then -(a-1) is a solution to the second. So b-1= -(a-1) that is a+b =2.
Adding them and completing the cube gave me:
(a+b)^3 - 3(a+b)^2 + 5(a+b) - 6 = 3ab(a+b) - 6ab
if a + b = u then:
u^3 - 3u^2 + 5u - 6 = 3ab(u - 2)
which led me to think: "if u = 2 then i get rid of the ab on the RHS, so let's test if u=2 is a solution"
in other words, I checked that value because it would be convenient for me to have a solution there
and when I checked it, I found it's a solution
not the most satisfying method, but it is proof that a+b = 2 is a valid solution
Thank you for your videos!
Awesome 😃👍.
Cumbersome and long method. Just by adding two equations and factorising would have got required answer.
Why not consider virtual roots?
my method: i added the two equations to get a³+b³-3(a²+b²)+5(a+b)-6=0 now i rewrites all terms with a+b and ab and got (a+b)((a+b)²-3ab)-3((a+b)²-2ab)+5(a+b)-6=0 and substituted a+b=u and ab=v and got finally u³-3u²+5u-3uv+6v-6=0 which easily factors in (u-2)(u²-u-3v+3)=0 from where u=2 but I is nothing but a+b=2 and the problem is that I don't know what to do with the second bracket.
Substitute u= a + b and v = ab in the second bracket. Then,
(a+b)^2 - ab - 3ab + 3 = 0
=> (a-b)^2 + 3 = 0
=> (a-b)^2 = -3
Which does not give real solution for all a,b belongs to R.
So, we can simply ignore this result.
This is what I think we should do with the other bracket
@@harshsamadhiya2326 Yeah it makes sense,thank you
Romanii exceleaza
@@dandeleanu3648 😅
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
If you complete the cube in the 1st and 2nd eqn you will get (a-1)^3 + 2(a-8)=0 and for the 2nd (b-1)^3 + 2(b+6)=0 if you add them you get (a-1)^3+(b-1)^3 +2(a-8) +2(b+6)=0, factor the sum of cubes and pull out a two and you get (a+b-2)(a^2 +b^2 -ab -a -b+3)=0 from this you get a+b=2 or a^2 +b^2 -ab -a -b +3=0 multiplying by two and completing the squares gives you, (a-b)^2 +(a-1)^2+(b-1)^2 +4=0, therefore no more real sol. for a,b.
Why did you solve it this way?
I found by calculs : a+b=-2 🤔
Wow
Can you plz solve this question a^1/3 +b^1/3 +c^1/3 = (2^1/3 - 1)^1/3 . Find a+b+c
There isn’t sufficient data I suppose,do you have a numerical ans ?
@@Soaring-Dragon do you use telegram i have this question on there. My friend send me this question
@@piyush-kx6vi I am sorry I can’t use telegram,try to know the numerical ans I suppose the question lacks some info and is based on transformation of roots.
쌌다