Question

Expand:
(i) ${\left(2x-\frac{2}{x}\right)}^3$
(ii) ${\left(3a+\frac{1}{4b}\right)}^3$
(iii) ${\left(\frac{4}{5}x-2\right)}^3$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(2x-\frac{2}{x}\right)}^3={\left(2x\right)}^3-{\left(\frac{2}{x}\right)}^3+3{\left(2x\right)}^2\left(-\frac{2}{x}\right)+3\left(2x\right){\left(-\frac{2}{x}\right)}^2 \\ =8x^3-\frac{8}{x^3}+3\left(4x^2\right)\left(-\frac{2}{x}\right)+3\left(2x\right)\left(\frac{4}{x^2}\right) \\ =8x^3-\frac{8}{x^3}+-24x+\frac{24}{x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right){\left(3a+\frac{1}{4b}\right)}^3={\left(3a\right)}^3+{\left(\frac{1}{4b}\right)}^3+3{\left(3a\right)}^2\left(\frac{1}{4b}\right)+3\left(3a\right){\left(\frac{1}{4b}\right)}^2 \\ =27a^3+\frac{1}{64b^3}+\frac{27a^2}{4b}+\frac{9a}{16b^2} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right){\left(\frac{4}{5}x-2\right)}^3={\left(\frac{4}{5}x\right)}^3-{\left(2\right)}^3-3{\left(\frac{4}{5}x\right)}^2\left(2\right)+3\left(\frac{4}{5}x\right){\left(2\right)}^2 \\ =\frac{64}{125}{x}^3-8-\frac{96}{25}{x}^2+\frac{48}{5}x \end{gathered}$$