Question

If $a\ne 0$ and $a-\frac{1}{a}=4$, find $a^4+\frac{1}{a^4}$.

• A. $92$
• B. $112$
• C. $322$
• D. $122$

Answer 1

The correct option is C $322$

$a-\frac{1}{a}=4$
Squaring both sides,
$=>(a-\frac{1}{a}{)}^2=4^2$
$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=16$
$=>a^2+\frac{1}{a^2}-2=16$
$=>a^2+\frac{1}{a^2}=16+2$
$=>a^2+\frac{1}{a^2}=18$
Again squaring both sides,
$=>(a^2+\frac{1}{a^2}{)}^2={18}^2$
$=>(a^2{)}^2+({\frac{1}{a}}^2{)}^2+2\left(a^2\right)\left(\frac{1}{a^2}\right)=324$
$=>(a^2{)}^2+(\frac{1}{a^2}{)}^2=324-2$
$=>a^4+\frac{1}{a^4}=322$