Question

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(5a-3b\right)}^3={\left(5a\right)}^3-{\left(3b\right)}^3-3{\left(5a\right)}^2\left(3b\right)+3\left(5a\right){\left(3b\right)}^2 \\ =125a^3-27b^3-225a^{2}b+135a{b}^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){\left(3x-\frac{5}{x}\right)}^3={\left(3x\right)}^3-{\left(\frac{5}{x}\right)}^3-3{\left(3x\right)}^2\left(\frac{5}{x}\right)+3\left(3x\right){\left(\frac{5}{x}\right)}^2 \\ =27x^3-\frac{125}{x^3}-135x+\frac{225}{x} \end{gathered}$$


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