Question

If $a\ne 0$ and $a-\frac{1}{a}=4$, then find $a^3-\frac{1}{a^3}$

• A. $76$
• B. $77$
• C. $75$
• D. $85$

Answer 1

The correct option is A $76$

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

Squaring both sides,

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

$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=16$

$=>a^2+\frac{1}{a^2}-2=16$

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

Now,

$=>a^3-\frac{1}{a^3}=(a-\frac{1}{a})\left[\right(a^2)+(\frac{1}{a}{)}^2+\left(a\right)\left(\frac{1}{a}\right)]$

$=>a^3-\frac{1}{a^3}=\left(4\right)(18+1)$

$=>a^3-\frac{1}{a^3}=76$