Question

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(3x+2\right)}^3={\left(3x\right)}^3+3\times {\left(3x\right)}^{2}x2+3\times 3x\times {\left(2\right)}^2+{\left(2\right)}^3 \\ =27x^3+54x^2+36x+8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right){\left(3a+\frac{1}{4b}\right)}^3={\left(3a\right)}^3+{\left(\frac{1}{4b}\right)}^3+3{\left(3a\right)}^2\left(\frac{1}{4b}\right)+3\left(3a\right){\left(\frac{1}{4b}\right)}^2 \\ =27a^3+\frac{1}{64b^3}+\frac{27a^2}{4b}+\frac{9a}{16b^2} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right){\left(1+\frac{2}{3}a\right)}^3={\left(\frac{2}{3}a\right)}^3+3\times {\left(\frac{2}{3}a\right)}^2\times 1+3a\frac{2}{3}a\times {\left(1\right)}^2+{\left(1\right)}^3 \\ =\frac{8}{27}{a}^3+\frac{4}{3}{a}^2+2a+1 \end{gathered}$$