Question

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

• A. $322$
• B. $321$
• C. $320$
• D. $319$

Answer 1

The correct option is A $322$

$a-\frac{1}{a}=4$

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,

$=>(a^2+\frac{1}{a^2}{)}^2={18}^2$

$=>(a^2{)}^2+({\frac{1}{a}}^2{)}^2+2\left(a^2\right)\left(\frac{1}{a^2}\right)=324$

$=>(a^2{)}^2+({\frac{1}{a}}^2{)}^2=324-2$

$=>(a^2{)}^2+({\frac{1}{a}}^2{)}^2=322$

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