Question

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

• A. $30$
• B. $32$
• C. $34$
• D. $36$

Answer 1

Answer: (D)

$36$

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

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,

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

$=3(a^2+1+\frac{1}{a^2})$

$=3(11+1)$

$=36$