Question

Find the following product:
$-\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^2-\frac{9}{4}x{y}^2{z}^3\right)$

Answer 1

To find the product, we will use the distributive law in the following way:

$$\begin{gathered} -\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^2-\frac{9}{4}x{y}^2{z}^3\right) \\ =\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{3}{2}xy{z}^2\right)\right\}-\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{9}{4}x{y}^2{z}^3\right)\right\} \\ =\left\{\left(-\frac{8}{27}\times \frac{3}{2}\right)\left(x\times x\right)\times \left(y\times y\right)\times \left(z\times z^2\right)\right\}-\left\{\left(-\frac{8}{27}\times \frac{9}{4}\right)\left(x\times x\right)\times \left(y\times y^2\right)\times \left(z\times z^3\right)\right\} \\ =\left\{\left(-\frac{8}{27}\times \frac{3}{2}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{8}{27}\times \frac{9}{4}\right)\left(x^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\} \\ =\left\{\left(-\frac{{\cancel{8}}^4}{{\cancel{27}}_9}\times \frac{\cancel{3}}{\cancel{2}}\right)\left(x^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{{\cancel{8}}^2}{{\cancel{27}}_3}\times \frac{\cancel{9}}{\cancel{4}}\right)\left(x^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\} \\ =-\frac{4}{9}{x}^2{y}^2{z}^3+\frac{2}{3}{x}^2{y}^3{z}^4 \end{gathered}$$

Thus, the answer is $-\frac{4}{9}{x}^2{y}^2{z}^3+\frac{2}{3}{x}^2{y}^3{z}^4$.