Question

If $x+\frac{1}{x}=4$, then $x^4+\frac{1}{x^4}=$?

Answer 1

Evaluate the given expression

Given equation, $x+\frac{1}{x}=4$

Square on both side,

${(x+\frac{1}{x})}^2=4^2$

By using identity ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$

$$\begin{gathered} \Rightarrow x^2+{\left(\frac{1}{x}\right)}^2+2\left(x\right)\left(\frac{1}{x}\right)=16 \\ \Rightarrow x^2+{\left(\frac{1}{x}\right)}^2+2=16 \\ \Rightarrow x^2+{\left(\frac{1}{x}\right)}^2=16-2 \\ \Rightarrow x^2+{\left(\frac{1}{x}\right)}^2=14 \end{gathered}$$

Again, squaring both sides, $\Rightarrow (x^2+{{\left(\frac{1}{x}\right)}^2)}^2=\left(14\right)$

$$\begin{gathered} \Rightarrow {\left(x^2\right)}^2+{\left(\frac{1}{x^2}\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)={14}^2 \\ \Rightarrow {\left(x^2\right)}^2+{\left(\frac{1}{x^2}\right)}^2+2=196 \\ \Rightarrow {\left(x^2\right)}^2+{\left(\frac{1}{x^2}\right)}^2=196-2 \\ \Rightarrow {\left(x^2\right)}^2+{\left(\frac{1}{x^2}\right)}^2=194 \end{gathered}$$

$\Rightarrow x^4+\frac{1}{x^4}=194$

Hence, $x^4+\frac{1}{x^4}=194$