Question

Simplify

(i) ${\left(\frac{3}{2}x-0.45y\right)}^2$

(ii) $\left(0.3x-0.2y\right)\left(0.9a-0.02b\right)$

(iii) $\left(0.3a-0.2b\right)\left(0.09a^2+0.06ab+0.04b^2\right)$

(iv) ${\left(0.7\right)}^3+{\left(0.3\right)}^3+3\times 0.7\times 0.3\left(0.7+0.3\right)$

(v) ${\left(3.5\right)}^3-{\left(1.5\right)}^3-3\times 3.5\times 1.5\left(3.5-1.5\right)$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(\frac{3}{2}x-0.45y\right)}^2 \\ ={\left(\frac{3}{2}x\right)}^2-2\times \frac{3}{2}x\times 0.45y+{\left(0.45y\right)}^2\left[{\left(a-b\right)}^2=a^2-2ab+b^2\right] \\ =\frac{9}{4}{x}^2-1.35xy+0.2025y^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\left(0.3x-0.2y\right)\left(0.9a-0.02b\right) \\ =0.3x\left(0.9a-0.02b\right)-0.2y\left(0.9a-0.02b\right) \\ =0.27ax-0.006bx-0.18ay+0.004by \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\left(0.3a-0.2b\right)\left(0.09a^2+0.06ab+0.04b^2\right) \\ =\left(0.3a-0.2b\right)\left[{\left(0.3a\right)}^2+\left(0.3a\right)\times \left(0.2b\right)+{\left(0.2b\right)}^2\right] \\ ={\left(0.3a\right)}^3-{\left(0.2b\right)}^3\left[a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ =0.027a^3-0.008b^3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right){\left(0.7\right)}^3+{\left(0.3\right)}^3+3\times 0.7\times 0.3\left(0.7+0.3\right) \\ ={\left(0.7+0.3\right)}^3\left[{\left(a+b\right)}^3=a^3+b^3+3ab\left(a+b\right)\right] \\ =1^3 \\ =1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right){\left(3.5\right)}^3-{\left(1.5\right)}^3-3\times 3.5\times 1.5\left(3.5-1.5\right) \\ ={\left(3.5-1.5\right)}^3\left[{\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right] \\ =2^3 \\ =8 \end{gathered}$$