Question

If $a^2+\frac{1}{a^2}=47$ and $a\ne 0$, find $a^3+\frac{1}{a^3}$ .

• A. $\pm 320$
• B. $\pm 322$
• C. $\pm 324$
• D. $\pm 328$

Answer 1

Answer: (B)

$\pm 322$

$a^2+\frac{1}{a^2}=47$

We know,

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

$=>(a+\frac{1}{a}{)}^2=47+2$

$=(a+\frac{1}{a}{)}^2=49$

$=(a+\frac{1}{a})=\pm \sqrt{49}$

$=>a+\frac{1}{a}=\pm 7$

Now,

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

$=>(\pm 7{)}^3=(a^3+\frac{1}{a^3})+3(\pm 7)$

If we take + sign we get,

$=>343=(a^3+\frac{1}{a^3})+21$

$=>a^3+\frac{1}{a^3}=322$

If we take - sign,we get,

$=>-343=(a^3+\frac{1}{a^3})-21$

$=>a^3+\frac{1}{a^3}=-322$

Thus,

$=>a^3+\frac{1}{a^3}=\pm 322$