Question

If $a\ne 0$ and $a-\frac{1}{a}=3$; find:

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

• A. $39$
• B. $36$
• C. $54$
• D. $14$

Answer 1

The correct option is B $36$

$a-\frac{1}{a}=3$
Squaring both sides,
$=>(a-\frac{1}{a}{)}^2=3^2$
$=>a^2+(\frac{1}{a}{)}^2-2(a\left)\right(\frac{1}{a})=9$
$=>a^2+\frac{1}{a^2}-2=9$
$=>a^2+\frac{1}{a^2}=11$
Now,
$a^3-\frac{1}{a^3}=(a-\frac{1}{a})\left[\right(a^2)+(a\left)\right(\frac{1}{a})+(\frac{1}{a^2}\left)\right]$
$=3(a^2+1+\frac{1}{a^2})$
$=3(11+1)$
$=36$