Ingenious Tricks for Evaluating the Hexic Algebraic Expression x^6+1/x^6 | Math Olympiad

This math challenge tutorial video teaches you a cool trick / method for finding the value of the algebraic expression x^6+1/x^6 given the value of x^4+1/x^4 using perfect square formula and sum of cubes formula.

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[nishant]

we could transpose 2 to left hand side in given equation and then it turns into (x²-1/x²)² = 0 which means x²= 1/x² which gives solution x²=1 which also gives answer to the question

[BlipBlop]

I learned from your previous video that one should start by working backward from the target expression for a while before proceeding forward with the given equation, so thanks for that!

[owlsmath]

Nice problem! Thanks! I was thinking... There is the property that when |x + 1/x| <= 2 then the values stay between -2 and 2 as exponent increases and pattern repeats. Working backwards in this case when x^4 + 1/x^4 = 2 then x + 1/x has 3 values: 0, -2, 2 but 0 means a non-real value for x. So must be -2 or 2 and those values mean that x^6 + 1/x^6 = 2. If we allow x + 1/x = 0 then we have a 2nd solution x^6 + 1/x^6 = -2

[piman]

The given equation reduces to x^4 = 1 => x = 1, -1, i. -i => x^6 = 1 or -1 => x^6 + 1/x^6 = 2 or -2.

[sy]

Thank you for explaining. Concerning the problem that uses "t+(1/t)=s", (in general, ) we use the method that we calculate [t+(1/t)]^2=s^2.
But, as for today's problem, I think other ways are easier and faster to solve:
[Method 1:] x^4+1/(x^4)=2 ∴ x^8+1=2×(x^4) ∴ x^8-2×(x^4)+1=0 ∴ (x^4-1)^2=0 ∴ x^4=1 ∴ (If x is a real number) x^2=1
∴ x^6+1/(x^6) = (x^2)^3+1/[(x^2)^3] = 1^3 + 1/(1^3) = 1+1 = 2
[Method 2:] (It can be used where x is a real number.) From the given equation, x≠0 ∴ x^4>0 ∴　x^4+1/(x^4)≧2 (∵ (A+B)/2≧√(AB), if A>0, B>0.)
x^4+1/(x^4)=2 ⇔ x^4=1/(x^4) ∴ x^8=1 ∴ x^4=1 ∴ x^2=1 ∴ x^6+1/(x^6) = (x^2)^3+1/[(x^2)^3] = 1^3 + 1/(1^3) = 1+1 = 2
By the way, does this problem require "x is a real number"?. If not, x may be a complex number. Today's video's title includes "Math Olympiad".
Compared to the level of normal Math Olympiad problems, if x is a real number, I think today's problem is too easy.
And if x is a complex number, the solution is not only "2". By using my "Method 1", x^4=1. ∴ x^2=±1 ( x = ±1, ±i )
If x is an imaginary number (that is i or -i), x^2=-1, and then, x^6+1/(x^6) = (x^2)^3+1/[(x^2)^3] = (-1)^3 + 1/[(-1)^3)] = (-1)+(-1) = -2
Therefore, the answer is " 2, -2 ". (: There are two solutions.)
Excuse me, but please check "in which books or documents the problem is written" and "what the problem condition is".

[razin]

plz mention that x is a real number in the begining of video