Question

If a ≠ 0 and ${\displaystyle a-\frac{1}{a}}$ = 4 ; find : ${\displaystyle (a^4+\frac{1}{a^4})}$

Answer 1

${\displaystyle {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}={(a-\frac{1}{a})}^2+2}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}={\left(4\right)}^2+2 [\because a-\frac{1}{a}=4]}$

⇒ ${\displaystyle a^2+\frac{1}{a^2}}$ = 18                        ...(1)

We know that,

${\displaystyle a^4+\frac{1}{a^4}={(a^2+\frac{1}{a^2})}^2-2}$

                       = `(18)^2 - 2             [ From(1) ]
                       = 324 - 2

⇒ ${\displaystyle a^4+\frac{1}{a^4}=322}$