If x^4 + x^(-4) = 7, what is the value of x^8 + x^(-8) =?

Given,
x^4 + x ^-4 = 7
(consider a = x^4 and b= x ^-4
if we square both the sides, (a+b)^2 = a*a + b*b + 2*a*b
here, 2*a*b will cancel out x term )
Squaring both the sides
x^8 + 2(x^8)(1/x^8)+1/x^8 = 49
x^8 + 1/x^8 = 47

Answer (C)

This is a simple question where you have to apply a basic algebraic identity to obtain the value of the given expression.
\((a+b)^2\) = \(a^2 + b^2 + 2ab\).
Therefore, \(a^2 + b^2 = (a+b)^2 – 2ab\)

\(x^4 + x^{(-4)}\) = 7, can be rewritten as \(x^4 + \frac{1}{x^4}\) = 7.

\(x^8 + x^{(-8)}\) = \(x^8 + \frac{1}{x^8}\). Observe that \(x^8 = (x^4)^2\); similarly, \(\frac{1}{x^8} = (\frac{1}{x^4})^2\).

So, we are essentially trying to find the value of an expression in the form of \(a^2 + b^2\).

Therefore, \(x^8 + \frac{1}{x^8}\) = \([x^4 + \frac{1}{x^4}]^2 – 2*x^4*\frac{1}{x^4} \)
= \([7]^2\) – 2 {since the \(x^4\) cancels out}
= 49 – 2 = 47.

The correct answer option is C.

Hope that helps!
Aravind B T

\(x^4 + \frac{1}{x^4} = 7\)

Squaring,
\(x^8 + \frac{1}{x^8} + 2 = 49\)

\(x^8 + \frac{1}{x^8} = 47\)

Hence, OA is C.

Can someone help me to know when to use certain algebraic tools? For example, here, how did you know that squaring both sides would help get the answer. To me, it seems as if it would be complicating it more, therefore making it more difficult, so my brain doesn't think it'll help. Thanks!

=> Let \(x^4 = A\) , then \(x^{-4} = \frac{1}{x^4} = \frac{1}{A}\)

=> \(x^4 + x^{-4} = 7 \)

=> \(A + \frac{1}{A} = 7 \)

=>\( A^2 + \frac{1}{A^2} + 2 = 49\) (On squaring both the sides)

=> \(A^2 + \frac{1}{A^2} = 49 - 2 = 47\)

=> \(x^8 = (x^4)^2 = A^2\)

=> \(x^8 + \frac{1}{(x^8)}= A^2 + \frac{1}{A^2} = 47\)

Answer C