Question

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(3x+2\right)}^3={\left(3x\right)}^3+3\times {\left(3x\right)}^{2}x2+3\times 3x\times {\left(2\right)}^2+{\left(2\right)}^3 \\ =27x^3+54x^2+36x+8 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){\left(3a-2b\right)}^3={\left(3a\right)}^3-3\times {\left(3a\right)}^2\times 2b+3\times 3a\times {\left(2b\right)}^2-{\left(2b\right)}^3 \\ =27a^3-54a^{2}b+36a{b}^2-8b^3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right){\left(\frac{2}{3}x+1\right)}^3={\left(\frac{2}{3}x\right)}^3+3\times {\left(\frac{2}{3}x\right)}^2\times 1+3x\frac{2}{3}x\times {\left(1\right)}^2+{\left(1\right)}^3 \\ =\frac{8}{27}{x}^3+\frac{4}{3}{x}^2+2x+1 \end{gathered}$$