Question

Find the following product:
$-\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^2\right)$

Answer 1

To find the product, we will use distributive law as follows:

$$\begin{gathered} -\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^2\right) \\ =\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{9}{2}{x}^{2}yz\right)\right\}-\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{3}{4}xy{z}^2\right)\right\} \\ =\left\{\left(-\frac{4}{27}\times \frac{9}{2}\right)\left(x^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{4}{27}\times \frac{3}{4}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\} \\ =\left\{\left(-\frac{{\cancel{4}}^2}{{\cancel{27}}_3}\times \frac{\cancel{9}}{\cancel{2}}\right)\left(x^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{{\cancel{4}}^1}{{\cancel{27}}_9}\times \frac{\cancel{3}}{\cancel{4}}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\} \\ =-\frac{2}{3}{x}^3{y}^2{z}^2+\frac{1}{9}{x}^2{y}^2{z}^3 \end{gathered}$$

Thus, the answer is $-\frac{2}{3}{x}^3{y}^2{z}^2+\frac{1}{9}{x}^2{y}^2{z}^3$.