Question

If $(a^2+\frac{1}{a^2})=14$; find the value of
$\left(ii\right)(a^3+\frac{1}{a^3})$

Answer 1

Given the value of Expression $(a^2+\frac{1}{a^2})=14$
Now, find the square of $(a+\frac{1}{a})$ by using the formula
$(a+b{)}^2=a^2+b^2+2ab$
$\Rightarrow {(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2\left(a\right)\left(\frac{1}{a}\right)$
$=14+2$
$=16$
$\therefore (a+\frac{1}{a})=\pm 4\cdots \left(1\right)$

Finding the value of $(a^3+\frac{1}{a^3})$

Now, find the cube of $(a+\frac{1}{a})$ by using the formula
$(a+b{)}^3=a^3+b^3+3ab(a+b)$

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

$\Rightarrow (\pm 4{)}^3=a^3+\frac{1}{a^3}+3(\pm 4)$ .....From[1]
$\therefore a^3+\frac{1}{a^3}=\pm (64-12)=\pm 52$
Hence, the value of $(a^3+\frac{1}{a^3})$ is $\pm 52$.