Question

Find: $a^3+\frac{1}{a^3}$, if $a+\frac{1}{a}=5$

Answer 1

Applying the formula,${(a+b)}^3=a^3+b^3+3ab(a+b)$

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

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

Substituting, $a+\frac{1}{a}=5$

$a^3+{\left(\frac{1}{a}\right)}^3=5^3-3\times 5=125-15=110$.