Question

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

• A. $666$
• B. $912$
• C. $672$
• D. $727$

Answer 1

The correct option is D $727$

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

Squaring both sides,

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

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

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

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

Again Squaring both sides,we get,

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

$=(x^2{)}^2+{\left(\frac{1}{x^2}\right)}^2+2(x^2)\left(\frac{1}{x^2}\right)=729$

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

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