Question

If $x-\frac{1}{x}=8$ find the value of $x^2+\frac{1}{x^2}$ and $x^4+\frac{1}{x^4}$

Answer 1

Step 1 : Identify the identity we can use, we can use ${(a-b)}^2=a^2-2ab+b^2$

If we square the term $x-\frac{1}{x}$ we get $x^2+\frac{1}{x^2}-2$, where we have the required terms

Step 2 : Find the value of $x^2+\frac{1}{x^2}$.

Squaring both sides of the given equation

$$\begin{gathered} {(x-\frac{1}{x})}^2=8^2 \\ {x}^2+\frac{1}{x^2}-2=64 \\ {x}^2+\frac{1}{x^2}=64+2 \\ {x}^2+\frac{1}{x^2}=66(\mathrm{equation}1) \end{gathered}$$

Step 3 : Find the value of $x^4+\frac{1}{x^4}$

Now squaring both side of the equation 1.

$$\begin{gathered} {(x^2+\frac{1}{x^2})}^2={\left(66\right)}^2 \\ {x}^4+\frac{1}{x^4}+2=4356 \\ {x}^4+\frac{1}{x^4}=4356-2 \\ {x}^4+\frac{1}{x^4}=4354 \end{gathered}$$

Hence the value of $x^2+\frac{1}{x^2}=66and{x}^4+\frac{1}{x^4}=4354$.