Question

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

$\left(\frac{4}{9}ab{c}^3\right)\times \left(\frac{-27}{5}{a}^3{b}^2\right)\times \left(-8b^{3}c\right)$

Answer 1

$$\begin{gathered} =(\frac{4}{9}\times \frac{-27}{5}\times -8)\times (\mathrm{a}\times {\mathrm{a}}^3\times \mathrm{b}\times {\mathrm{b}}^2\times {\mathrm{b}}^3\times {\mathrm{c}}^3\times \mathrm{c}) \\ =\frac{96}{5}\times {\mathrm{a}}^{(1+3)}\times {\mathrm{b}}^{(1+2+3)}\times {\mathrm{c}}^{(3+1)} \\ =\frac{96}{5}{\mathrm{a}}^4{\mathrm{b}}^6{\mathrm{c}}^4 \\ \mathrm{When}\mathrm{a}=1,\mathrm{b}=2\mathrm{and}\mathrm{c}=3: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:(\frac{4}{9}\times \frac{-27}{5}\times -8)\times (\mathrm{a}\times {\mathrm{a}}^3\times \mathrm{b}\times {\mathrm{b}}^2\times {\mathrm{b}}^3\times {\mathrm{c}}^3\times \mathrm{c}) \\ =(\frac{4}{9}\times \frac{-27}{5}\times -8)\times \left(1\times 1^3\times 2\times 2^2\times 2^3\times 3^3\times 3\right) \\ =\frac{497664}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{96}{5}{\mathrm{a}}^4{\mathrm{b}}^6{\mathrm{c}}^4=\frac{96}{5}\left(1^4\times 2^6\times 3^4\right)=\frac{497664}{5} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence},\mathrm{verified}. \end{gathered}$$