Question

If $2x-\frac{1}{2x}=4$, find $8x^3+\frac{1}{8x^3}$.

Answer 1

Given, $2x-\frac{1}{2x}=4$

Cubing both sides,
${(2x-\frac{1}{2x})}^3=64$
Applying the formula,

${(a-b)}^3=a^3-b^3-3ab(a-b)$

$\Rightarrow {\left(2x\right)}^3-{\left(\frac{1}{2x}\right)}^3-3\times 2x\times \frac{1}{2x}\times (2x-\frac{1}{2x})=64$

$\Rightarrow 8x^3-\frac{1}{8x^3}-3\times 4=64$ [$\because 2x-\frac{1}{2x}=4$]

$\Rightarrow 8x^3-\frac{1}{8x^3}=64+12=76$

Hence, $8x^3+\frac{1}{8x^3}=76$