Question

If $x-\frac{1}{x}=5,$ then find the value of $x^4+\frac{1}{x^4}$.

Answer 1

Given, $x-\frac{1}{x}=5$

Squaring both sides,

${(x-\frac{1}{x})}^2=5^2$

${(x-\frac{1}{x})}^2=25$

$\Rightarrow x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}=25$ .......... $(a-b{)}^2=a^2-2ab+b^2$

$\Rightarrow x^2+\frac{1}{x^2}-2=25$

$\Rightarrow x^2+\frac{1}{x^2}=25+2=27$

$\therefore x^2+\frac{1}{x^2}=27$

Squaring both sides,

${(x^2+\frac{1}{x^2})}^2={27}^2$

$x^4+\frac{1}{x^4}+2\times x^2\times \frac{1}{x^2}=729$ ............ $(a+b{)}^2=a^2+2ab+b^2$

$x^4+\frac{1}{x^4}+2=729$

$x^4+\frac{1}{x^4}=729-2=727$

$\therefore x^4+\frac{1}{x^4}=727$