Question

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

$\left(\frac{1}{4}abc\right)\times \left(-6b^{2}c\right)\times \left(-\frac{1}{3}{c}^3\right)$

Answer 1

$$\begin{gathered} =(\frac{1}{4}\times -6\times -\frac{1}{3})\times (\mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c}\times \mathrm{c}\times {\mathrm{c}}^3) \\ =\frac{1}{2}\times \mathrm{a}\times {\mathrm{b}}^{(1+2)}\times {\mathrm{c}}^{(1+1+3)} \\ =\frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5 \\ \mathrm{When}\mathrm{a}=1,\mathrm{b}=2\mathrm{and}\mathrm{c}=3,\mathrm{we}\mathrm{get}: \\ \frac{1}{4}\mathrm{abc}=\frac{1}{4}\times 1\times 2\times 3=\frac{3}{2} \\ -6{\mathrm{b}}^2\mathrm{c}=-6\times 2^2\times 3=-72 \\ -\frac{1}{3}{\mathrm{c}}^3=\frac{-1}{3}\times 3^3=-9 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=\frac{1}{4}\mathrm{abc}\times -6{\mathrm{b}}^2\mathrm{c}\times -\frac{1}{3}{\mathrm{c}}^3=\frac{3}{2}\times -72\times -9=972 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5=\frac{1}{2}\times 1\times 2^3\times 3^5=972 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \frac{1}{4}\mathrm{abc}\times -6{\mathrm{b}}^2\mathrm{c}\times -\frac{1}{3}{\mathrm{c}}^3=\frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5 \end{gathered}$$