Question

Find: $a^3-\frac{1}{a^3}$, if $a-\frac{1}{a}=4$

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, $a-\frac{1}{a}=4$

$a^3-{\left(\frac{1}{a}\right)}^3=4^3+3\times 4=64+12=76$.