Can you solve this equation?
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- Опубліковано 28 чер 2024
- This is an interesting question that tests a lot of concepts!
0:00 problem
0:38 review formulas
4:28 solution
@drpkmath1234
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Wow, thanks for mentioning me Presh. Thanks for solving this problem so beautifully. Kudos to you!
Glad to see ya here prof.
I think there's a simpler solution.
Let: x = y³
Square both sides: 1 + √(1+y³) = y²
Minus 1 and square: 1 + y³ = y⁴ - 2y² + 1
Rearrange and factor: 0 = y⁴ - y³ - 2y²
Factorise: 0 = y²(y² - y - 2)
Factorise: 0 = y²(y - 2)(y + 1)
Substitute: x = 0, x = 8, x = -1
Check: x = 8 is the only solution.
hey I did this in the same way too
technically all of them work if you consider sqrt(1)=+ -1
@@abhijiths5237 There'd be other solutions too if we considered 1+1=3.
However, since 1+1≠3 and √1≠-1, those solutions don't work.
@@beng4186 I'm talking about the multivariable roots. Let √x=1. So x²=1 and x can be +1 or -1 as -1² also equals 1. We consider square roots generally as +ve so most of the time we ignore -1. Haven't you heard of the cube roots of unity?
interestingly its only one solution, x=8, and no complex once
for fun tested √(1+√(1±x) )= (±x)¹/ ³
..the solutions is 8, -8, 1, -1 ...
∛is cubroot symbol (the YT font does a poor job displaying it clearly), using ¹/ ³ instead
I used a substitution for √(1 + x). This makes x = m^2 - 1. Sub in and raise both sides to the 6th power. Since m + 1 =\= 0, divide both sides by (m + 1)^2. Simple after that. -1 and 8, with -1 extraneous
Yup, that's exactly the way that I did it.
@@philstubblefield Same here. Instant simplification. No more nested root. Kinda obvious.
same here
Same
It's way simpler without utilizing the difference of squares or sum of cubes. The term 1 on both sides just cancels out, then you can factorize the polynomial and get smth like u^2(u^2-u-2). The quadratic is easily factorizable by guess to (u-2)(u+1).
Same for me. I solved it and then watched the video wondering the whole time WHY was he making it so complicated?!?
(Probably to throw in the formulas as a bonus lesson, I guess...)
The reason you cannot do what you suggest is that your solution requires you to divide both sides by u + 1. There is no guarantee that u -1, which results in dividing by zero.
@@salkabalani1482 no. If you just move everything to one side, you can factor without any division. And even if there was a division, we would just have to check u=1 separately.
Instead of any substitution, I did it by raising LHS and RHS to the power of 6 (LCM of 2 and 3), then ending with an equation containing only √(x+1) and other x terms, which then can be squared to get a polynomial equation, and solve them... It takes some steps to solve, but hey, its a solution...
If someone aa curious as me, there are always some ways to make extraneous solutions work.
x=-1 is solution if we assume external square root as a branch of multivalued function fixed by sqrt(1) = -1
x=0 is solution if internal square root is a branch fixed by the same thing as above.
By inspection X=8
I did that too, but it didn't clearly mean it was the only solution without a little analysis -- I looked at neighborhoods of the solution to convince myself, it was the only one.
Lmao I did this too!
It's easier than that: squaring both sides you get (1 + x)^(1/2) + 1 = x^(2/3). Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3). Subtracting the two equations gives x^(2/3) - x^(1/3) = 2. Letting y = x^(1/3) we get that y^2 - y - 2 = 0, which has solutions y = 2 and y = -1, corresponding to x = 8 and x = -1 respectively. Only the first fits the original equation, so x = 8 is the only solution.
"Since the product of (1 + x)^(1/2) + 1 and (1 + x)^(1/2) - 1 is x, we have that (1 + x)^(1/2) - 1 = x^(1/3)"
How did you get to that from that product?
@@maxhagenauer24 We are using that if ab = x and a = x^(2/3), then b = x/a = x/(x^(2/3)) = x^(1/3).
Where do you get the equation x^(2/3) - x^(1/2) = 2?
@@maxhagenauer24fooled me for a moment too:
Since the first equation gave us
1+sqrt(1+x) = x^2/3
[1+sqrt(1+x)] * x^1/3 = x
Now noting that
[sqrt(1+x)+1] [sqrt(1+x)-1] = x
Setting them equal we can divide by the like term and get
x^1/3 = sqrt(1+x) -1
@@ericanibal6222 Subtract the equation (1 + x)^(1/2) - 1 = x^(1/3) from the equation (1 + x)^(1/2) + 1 = x^(2/3)
u=1+sqrt(1+x) is a way nicer substitution, for anyone who wants to try
I wish you had more videos on stereometry, Presh! It's very enlightening.
Basic field theory says that the cube root of x cannot be expressed as any combination of square roots unless x is a perfect cube. (This is why doubling the cube with only a straightedge and compass is an impossible construction.) From there, simple inspection gives x = 8.
This was Finnish math olympiad problem Dr PK Math posted about a month ago.
wow! i really like that visual explanation for the formulas, it really helped me understand!
1st! Really love your videos! They always help me to learn something new. Thanks ❤
Incredible. Another great one.
I had to pause the video at points but I understood the method. I'm glad you mentioned not to stop since after that long calculation I would have chosen all three numbers to be the final answer which is wrong.
I did the calculation in the old-fashioned way, entirely without substitution. I raised both sides to the sixth power. (You can also first square both sides and then cube them both, if you prefer that.)
Then, I isolated the term with the root and squared the equation again.
One might assume that this results in a super unwieldy equation but after simplifying the terms, one is left with this:
7x³ + 8x² = x⁴
We can further simplify by dividing by x² (and testing whether x²=0 and therefore x=0 is a solution; it is of course not one), and we get this simple quadratic equation:
x² - 7x - 8 = 0
This equation has two solutions, x=-1 and x=8; testing shows that x=-1 is an extraneous solution but x=8 is valid.
And that's it! 😀
Love this solution and graphical representation of those formulae ❤
Love this stuff
I loved the geometric representation at the beginning of the explanation. Makes it so much easier to visualize where these algebraic equations originated.
A cool use for the difference and sum of cubes formulas in my opinion is rationalising the denominator when it has a cube root
I just saw x = 8 by inspection and I did some calculus reasoning to find that there probably aren't any more solutions
Me too!
I did this with way less steps and not needing difference of cubes or squares. I just didn't do a u substitution and worked on getting rid of the squares, followed by some simplification and factoring. I was able to get the same answers in about about 9 lines of work. I do like that you can take advantage of the difference of cubes and squares if you want.
Very good video. Thanks
Just from the expected shapes of the functions I would only expect one solution if there is any. You have a cube root and something that will generally act like a fourth root. Those two only intersect at 0 and 1, but the adding of 1s will shift it around so I don’t expect them to meet up at both of those two spots anymore.
Not a solution, but the quick analysis so I can see if what I get in the end makes sense.
Do the u substitution and rearrange till you get -u^4 + u^3 + 2u^2 = 0, factor out u^2 and solve for u to be 0, 2, and -1. Sub these back in to find x and you get that the only one that works is 8.
I absolutely loved the visual explanation of the diff of squares and cubes! While I agree with many of the people here about the simpler way of doing it. I did the same all the while wondering if I was losing any mathematical rigor in the process (I have learned that you can’t cancel things willy nilly even tho’ it may be obvious). am curious to know why you did not go that route.
Answer 3
Explanation:
Square both sides first
Then
Solve using substitute √1+x = t then substitute x= t^2 - 1 and make equation in 't' then solve for t quadratic equation then t=2 and x will equal to' 3'
So 3 is the answer
It is possible to dispense with the extraneous solutions at the very start by noticing that the left-hand side of the original equation can't have a value of less than 1 for every possible value of x in the domain of that formula. This means that cbrt(x) can't be less than 1; that is, x can't be less than 1. Since x=8 is the only solution that fits this criterion, it's the only solution of the equation.
I substituted x for a^3 and it worked really easy
That is the most roundabout and incoherent solution video I've ever seen. This is what I did:
Square both sides and rearrange to get: sqrt(1+x) = x^(2/3) - 1
Square both sides again to get: 1+x = x^(4/3) - 2x^(2/3) + 1
Simplification yields: x^(4/3) - 2x^(2/3) - x = 0
Now make the following substitution: u = x^1/3
u^4 - u^3 - 2u^2 = 0 It's obvious that you can factor out u^2 here.
u^2 (u^2 - u - 2) = 0 u=0 is a root.
Solving the quadratic in parenthesis yields: u = (-1,2)
Now back-substitute to convert u's to x's. x = u^3
x = (0^3, (-1)^3, 2^3) = (0,-1,8)
Check the above values for x against the original problem equation to determine actual solution(s) and discard any extraneous roots.
When running that check, only one solution emerges: x = 8. Done.
square both sides, subtract 1 then square both sides again. set u = x^1/3 and it's an easy solution
Beautiful.. thanks..
Nice sum
I did this doing two substitutions, x = a^2-1, then a = b^2-1, then cubed both sides and solved to get b = -1,0,2 going back through the substitution chain, I get x = -1, 0, 8. -1 and 0 are extraneous, leaving 8.
Fantastic
Why not get to x^(4/3)-2x^(2/3)= x and substitute x=y^3. y^4-2y^2=y^3. Check for y=0 and then solve y^2-y=2. y=2 or -1 and x=8 or -1 but x=-1 is invalid.
yeah thats what i did lol
I solved it but I think I made it much harder for myself. I didn't substitute at all and only brute forced it by cubing and squaring the entire equation and distributing the terms whenever possible. At one point I had to distribute two terms both with 3 different terms and thus ended up with 9 terms. I was surprised that in the end I was left with a quadratic equation that had a really simple discriminant. And after checking the solutions, those being -1 and 8, 8 turned out to be the correct solution. I'd like to know if anyone else went about the problem this way or if it's just a needlessly complicated process
I raised both to the sixth power, then eventually got to:
(x+4)(x+1)^0.5 = x^2-3x-4
Squared both sides of that and got:
x^4-7x^3-8x^2=0
This factors to (x-8)(x+1)(x^2)=0. The rest follows the same way as 8:55 on.
X=8 at first glance 😂
But were you sure it was the only solution?
@@Kero-zc5tc just check cube numbers as in most of the case answer comes out to be the simplest one
Seriously!
Wolphram alpha, Mathway and similar apps can solve it very nicely.
@@himanshuuu6361 I think you missed my point, I was saying you don’t know if 8 is the only solution I was not at all referring to how to guess a solution
I solved this in a simpler manner, I think.
I used the same u substitution and squared both sides, getting sqrt(1 - u^3) = u^2 - 1. I squared both sides again to get 1 + u^3 = u^4 - 2u^2 + 1. The ones cancel, and moving everything to the left we have u^4 - u^3 - 2u^2 = 0. This factors to (u^2)(u^2 - u - 2) = 0
Note that u cannot be negative, as the cube root of x is a principal square root. Also note that u cannot be 0 because we can test it in the original equation and show it will not work. Therefore u is strictly positive.
Since u cannot be zero, we can divide both sides by u^2 with no worries, leaving is with the quadratic which is easy and yields u = 2 or -1. u is strictly positive, so u = 2, and since x = u^3, x = 8.
It literally took me 5 seconds to solve this in my head by inspection!
The answer is 8, simply done by trying trying out numbers that gave an integer answer on the left. So sqrt(x+1) must be integer. X=0 doesn't work, as we get sqrt(2) on the left, but the next candidate x=3 gives 2 on the left and the right. Of course, if the answer wasn't integer, this wouldn't work, but if time was critical in a test, worth a few seconds to try it.
As others have pointed out, this solution is not the simplest. Here's what I did:
(1 + (1 + x)^(1/2))^(1/2) = x^(1/3)
1 + (1 + x)^(1/2) = x^(2/3) -Square both sides
(1 + x)^(1/2) = x^(2/3) - 1 -Move 1 to the other side
1 + x = x^(4/3) - 2x^(2/3) + 1 -Square both sides
x^(4/3) - x - 2x^(2/3) = 0 -Move everything to one side (1's cancel)
[x^(2/3)][x^(2/3) - x^(1/3) - 2] = 0 -Factor out x^(2/3)
[x^(2/3)][x^(1/3) - 2][x^(1/3) + 1] = 0 -Factor the quadratic in terms of x^(1/3)
Then each factor can be set equal to 0 and solved for x. The same solutions of 0, 8, and -1 emerge. Only 8 works.
Does an extraneous solution mean an impossible solution?
extraneous solution means solutions obtained when we square or cube both sides but those solutions aren't actually true.
@asr2009 has the right of it. In general, extraneous solutions are introduced by the math used to solve the problem. For radical equations, extraneous solutions can occur when you raise the equation to an even power (this is because doing so destroys information about signs).
For a different type of extraneous solution, rational equations (where there is a variable in the denominator) can have extraneous solutions introduced by multiplying. When you plug those in to check, they generally cause division by zero.
In all cases, extraneous solutions are solutions that do not work in the original equation, but only in the transformed one.
@@davidtripp2118thanks!!
@@davidtripp2118 When I looked up the graphs of *√(1+√(1+x)) = x⅓* and it's transformed equation *x⁴ - 7x³ - 8x² = 0* on Desmos, It showed X = 8 for first equation and X = -1 & 8 for the second equation.
We can clearly see that X = 0 is also a solution for the second equation, so why does it eliminate that possibility?
The extra solutions only "work" if you take the negative square roots of the original equation -- not something done in formal mathematics!
For example, for x = 0 you can take the negative square root of the inner sqrt(1 + x) term to get -1, then add that to the 1 and get sqrt(0) = 0^1/3 = 0.
Also for x = -1, the inner sqrt term is zero (1 + -1), resulting in sqrt(1) = (-1)^1/3. Take the negative sqrt of 1 there to get -1 = -1.
It's issues like these that caused mathematicians to regulate using only the "principle square root" in equations, which is why the 0 and -1 solutions don't work in practice.
Let's elevate each side to the 5th power:
( 1 + √(1 + x) )^3 = x²
Setting t=√(1 + x), we get the system:
{ x² = (1+t)^3
{ t² = x+1
=> t²-1 = x
=> (t-1)²(t+1)² = x² = (1+t)^3
=> (t+1)² (t+1+(t-1)²) = 0
=> (t+1)².t.(t-3) = 0
t = 3 is the only valid solution
=> x = 8
Can someone explain to me if, and if so, how, we can make sure the value we got is true without the need of substitution?
Retrace your steps and do them in reverse, if there is no problem then all the solutions should work.
If for example I square both sides of an equation, and then try to go back (using square root) that would give me the absolute values of both sides, which is not what I started with (unless I know that they are positive for sure), so I can conclude that some of the solutions I find might not work
I have two questions... How do we know that that this is in fact a quartic (p=4)? And I got the same extraneous solutions as well. Where do the other three extraneous solutions come from? (x=-1 and x=0, twice)
Edit: Fwiw I raised to the 6th first, did some algebra, and then squared to get the quartic. The quartic was super easy since the x^1 and x^0 terms were 0.
Simple try to insert a couple of third power numbers give us 8 in 20 seconds or so, why so long way so solve?
This shows why you start solving such problems by trying out a few simple numbers. Right hand side has a cube root of x, so you try the first cube numbers: 0: nope, 1: nope, 8: bingo!
Well, that is only true if you assume there's going to be integer solutions. Of course it doesn't take long to try first few integers.
By Mihailescu's theorem, the only perfect powers that differ by 1 are 8 and 9. So there's only one candidate to check.
Please tackle this one 🙏🙏
if a= √3 +√4 +√5 find a⁴ - 8a³ +8a² +32a
Use chatgpt
I got the same 3 solutions right (in a different way: squaring the roots, and then I made y=3✓x; I solved for y, and then found the 3 values for x)
But I didn't check if that was right substituting in the original equation😔
Hey, please read this, I got this problem stuck in my head for the past month. If I draw a rectangle and on the long side of it I draw a square with the same dimensions as the long side and then I lengthen the side to infinity, the area of the square is infinity but also the one of the rectangle but the squares infinity is bigger than the infinity of the rectangle, why?
sqr(1) could be -1, then, the fist option could be a valid real solution.
try substituting -1 and 0 to the original equation and see if its true. (8 is true)
Domain of the function was x>0 anyway, need not check for x = -1, 0
I don't know why all solutions presented are so ridiculously complicated. All one has to do is square both sides, rearrange, square both side again, cancel out the 1 on both sides, divide by x, substitute y=x^(1/3) , solve the quadratic and cube the root to find x.
cbr of negative one is NOT negative one! We're talking about principle values, not real valued roots.
Damn it was deep
I did it in my head in about 20 - 30 seconds by substituting likley numbers.
Square both side to solve algebrically
Omg I graduated high school without knowing where the formula for the diff of square and cube came from! This came thirteen years too late for me. 😂
9:48 why here -0? I know that +0 and -0 is same things but equation was √(1+√(1+x))=³√x and not √(1+√(1-x))=³√x
I did the following:
Raise to the power of 6 (since the LCM of 2 and 3 is 6) on both sides:
(1 + √(1+x))³ = x²
=> 1 +3(1 + x) + 3 √(1 + x) + (1+x)√(1+x) = x²
=> (4 + 3x) + (x + 4)√(1 + x)=x²
=> (x + 4)√(1 + x) = x²-3x-4
=> (x + 4)√(1 + x) = (x+1)(x-4) or x + 1 = 0 => x = - 1
=> x + 4 = (x-4)√(1 + x)
=> (x+4)² = (x-4)²(1+x)
=> (x² - 8x + 16) (1+x) - (x² + 8x + 16) = 0
=> x² - 8x + 16 + x³ - 8x² + 16x - x² - 8x - 16 = 0
=> x³ - 8x² = 0
=> x = 0 or x = 8
Check x = - 1 is not a solution, since the primary square root of 1 = 1, not -1.
Check x = 0 is not a solution, since the right-hand side is 0 in the original equation, while the left is √2.
So the only solution is x = 8.
i tried the same but only made it to the second line and gave up. didnt think of moving 4+3x to the rhs
Given
√{1+√(1+x)} = x^(1/3)
Squaring both side
1+√(1+x)=x^(2/3)
Rearranging the terms
1-x^(2/3) = √(1+x)
Squaring both side
1 + x^(4/3) - 2x^(2/3) = 1+x
Rearranging and cancelling terms
x^(4/3) -2x^(2/3) - x^(3/3) = 0
Taking ∛x as 'm'
so ⇒ m⁴ - 2m² - m³ = 0
Taking m² common
m²(m²-m-2) = 0
m²(m-2)(m+1) = 0
So m=0 or -1 or 2
⇒ ∛x = 0 or -1 or 2
⇒x= 0 or -1 or 8
Putting x=0
⇒1=0 which is not possible
Putting x = -1
⇒1 = ∛(-1) Which is not possible
Putting x=8
⇒2=2
*So x=2 is the only solution*
But I have a doubt, if x=2 is the only solution then why we got 3 solution in first place ??
TLDR: brute-forcing this problem is not that hard, but this video's solution looks much cooler than mine.
Here's how I've done it, by cubing both sides once and squaring both sides twice in total: you can see how I exploited the 'x must be real' condition to its limits. First I cubed and squared both sides. The left-hand side becomes (1 + sqrt(1 + x))³, the right-hand side becomes x². I unpacked the left-hand side and transferred all elements without the sqrt(1+x) term to the right-hand side. The resulting equation is: (4+x)*sqrt(1+x) = x² - 3x - 4. I modified the right-hand side into (x-4)(x+1). *x should be real number* so x >= -1, but when x=-1 the equation does not work. So x > -1. Square both sides, divide both sides with (x+1), and when I transferred all terms to one side I ended up with x³ - 8x = 0, or *x²(x-8) = 0.*
x is 8.
Wow! I pretty much guessed that x could be 8, and I was right. I did none of that extra math.
Solved it by guessing :D
X=8 my friend by inspection
I solved by substituting root(x+1) with t and then after solving for t. I got x=8 as the answer directly😅.
at first glance, I immediately squared both sides to get 1+sqrt(1+x) = x^(2/3)
moving the +1 from LHS to RHS, sqrt(1+x) = x^(2/3) - 1
squaring both sides again, 1 + x = x^(4/3) - 2*1*x^(2/3) + 1
since there is +1 on both sides, they can cancel out
so we get, x = x^(4/3) - 2*x^(2/3)
moving the x from LHS to RHS and switching the sides(since we can do that)
x^(4/3) - 2*x^(2/3) - x = 0
x^{1+(1/3)} - 2*x^{1-(1/3)} - x = 0
using exponent rules, it implies that,
x*x^(1/3) -2*x*x^(-1/3) - x = 0
x{x^(1/3) - 2*x^(-1/3) - 1} = 0
so, x=0 seems like a solution here, but the second part of the equation makes absolutely no sense with x=0 and as seen in the video, it does not work.
then, we say the second multiple of the LHS is equal to zero for the second case
just for ease of understanding, I'll suppose t=x^(1/3)
then the second multiple becomes:
t - 2/t - 1 = 0
since t is not equal to 0, I can multiply t on both sides
t^2 - 2 - t = 0
t^2 - t - 2= 0 [rearranging]
t^2 - 2t + t - 2 = 0
t(t-2) + 1(t-2) = 0
(t+1)(t-2)=0
then, for the second case(since we had a case of x=0), we get t+1=0
then t=-1
x^1/3 = -1
I do not need to put it back in to know that it will not work. That is because x^1/3 is the RHS of the original equation and it CANNOT be negative since there is a square root on the LHS which prevents it from being negative.
so the final case is t-2=0
t=2
x^(1/3)=2
x=8[by cubing both sides]
therefore, x=8
Why cant root of 1 be (-1)?
Square root of 1 can also be -1 so why isn't X = -1 a valid solution?
Is it just me, or does everyone glance at the length of the video to estimate the difficulty of the problem? : )
Nope, not just you.
Just trying the first few easy integers, the answer is 8. But I doubt that’s what was meant by “solve the equation”.
I wish
❤❤❤❤❤
x=8
For me is strange why solving the equation we found a solution inside the domain of the x (x=0) and suddenly this solution is not right...
You forgot the negative square root of 1.
Trial and error
My first guess was x=8😅
Put x=tan²y
Than it would be easy to solve
How some x cant work with the original equation?
I wonder ff it took me 1 minute to basically guess that x=8 just by trying it out with pen and paper (and I'm as bad at math as it gets) then I'm not sure if it's just luck or can you really just guess...?
8
[ sqrt 1 + sqrt ( 1+x)]^6 = (cube root of 3)^6 raised both sides to the 6th power
[1 + sqrt ( 1 + x ]^3 = x^2
let sqrt (1 + x) = n
then 1 + x = n^2
and x = n^2 -1
Hence, (1 + n)^3 = (n^2 -1)^2
1 + n^3 + 3n + 3n^2= n^4 -2n^2 + 1
0 = n^4- n^3 -5n^2 - 3n
0 = n^3 - n^2 -5n -3 (divide both sides by n
0= n^3 -3n - n^2 - 2n -3
0 = n^3- 3n- (n-3)(n+1)
0 = (n+1) * (n-3)n+1)
n=-1 n=-1 and n=3
Hence, sqrt (1+ x) =3
1+ x =9
x=8 Answer
sqrt (1+ x) also = -1
1+ x =1
x=0 but 0 does not satisfy the equation.
the substitution thing feels like cheating
i got (7+-9)/2
I graphed it and came up with 8.
yes. i can.
Chat gpt takes 10 seconds to solve with all possible solutions
Why are there wrong solutions for x?
I didn't pause :)
X=-1 works if you take the negative square root when you calculate sqrt(1). X=0 also works if you take the negative square root when you calculate sqrt(1-0).
However, you're *NOT* allowed to do that, so you should not bring that up.
Why not? It’s a solution to the square root?
So long since I had to do that stuff (school, decades ago), but that's elegant, in my opinion. Thx!
Yes. His strategy is to do the substitution from the beginning. That might be useful.
I just looked at the equation and in 2-3 seconds my gut tells me x=8
Why oh why use geometry???
No I can't and explain to me why I should want to
Easy, if you see the trick. X=8. Did it by looking. (Got lucky, I think.)
I just saw that root (root 9 + 1) = 2 = cube root of 8
x = 8 Thats it 😂😂😂😂
I did Whole square both sides then I got √(1+x)=x⅔-1
Then again whole square 😂 I got 1+x =x^(4/3)+1-2x⅔
Then +1 is cancelled so we get x=x^(4/3) - 2x⅔
Then take x common from RHS
To get x=x{x⅓-2x^(-1/3)
So x cancelled x
We get 1=x⅓-2x^(-1/3)
Put x⅓=T (just a variable)
So we get 1=T-2/T
Here form a quadratic T²-T-2=0
(U BETTER KNOW HO TO SOLVE THIS🗿)
T=2 THERE FORE X⅓=2
X=2³=8
IF U READ THIS LONG Thanks IT TOOK ME 6 MIN 40 SEC TO TYPE 😂👍👍👍👍