Question

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

$a^3-\frac{1}{a^3}$

• A. $76$
• B. $71$
• C. $45$
• D. $16$

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,
$=>a^3-\frac{1}{a^3}=(a-\frac{1}{a})\left[\right(a^2)+(\frac{1}{a}{)}^2+\left(a\right)\left(\frac{1}{a}\right)]$
$=>a^3-\frac{1}{a^3}=(a-\frac{1}{a})\left[\right(a^2)+(\frac{1}{a}{)}^2+1]$
$=>a^3-\frac{1}{a^3}=\left(4\right)\left(19\right)$
$=>a^3-\frac{1}{a^3}=76$