Question

Find the product and verify the result for a = 1, b = 2 and c = 3.

$\left(\frac{2}{5}{a}^{2}b\right)\times \left(-15b^{2}ac\right)\times \left(-\frac{1}{2}{c}^2\right)$

Answer 1

$$\begin{gathered} =(\frac{2}{5}\times -15\times \frac{-1}{2})\times ({\mathrm{a}}^2\times \mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c}\times {\mathrm{c}}^2) \\ =3\times {\mathrm{a}}^{(2+1)}\times {\mathrm{b}}^{(1+2)}\times {\mathrm{c}}^{(1+2)} \\ =3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \\ \mathrm{When}\mathrm{a}=1,\mathrm{b}=2\mathrm{and}\mathrm{c}=3,\mathrm{we}\mathrm{get}: \\ \frac{2}{5}{\mathrm{a}}^2\mathrm{b}=\frac{2}{5}\times 1^2\times 2=\frac{4}{5} \\ -15{\mathrm{b}}^2\mathrm{ac}=-15\times 2^2\times 1\times 3=-180 \\ \frac{-1}{2}{\mathrm{c}}^2=\frac{-1}{2}\times 3^2=\frac{-9}{2} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=\frac{2}{5}{\mathrm{a}}^2\mathrm{b}\times -15{\mathrm{b}}^2\mathrm{ac}\times \frac{-1}{2}{\mathrm{c}}^2=\frac{4}{5}\times -180\times \frac{-9}{2}=648 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3=3\times 1^3\times 2^3\times 3^3=648 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \frac{2}{5}{\mathrm{a}}^2\mathrm{b}\times -15{\mathrm{b}}^2\mathrm{ac}\times \frac{-1}{2}{\mathrm{c}}^2=3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \end{gathered}$$