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:

\[- \frac{4}{27}xyz\left( \frac{9}{2} x^2 yz - \frac{3}{4}xy z^2 \right)\]

\[ = \left\{ \left( - \frac{4}{27}xyz \right)\left( \frac{9}{2} x^2 yz \right) \right\} - \left\{ \left( - \frac{4}{27}xyz \right)\left( \frac{3}{4}xy z^2 \right) \right\}\]

\[ = \left\{ \left( - \frac{4}{27} \times \frac{9}{2} \right)\left( x^{1 + 2} y^{1 + 1} z^{1 + 1} \right) \right\} - \left\{ \left( - \frac{4}{27} \times \frac{3}{4} \right)\left( x^{1 + 1} y^{1 + 1} z^{1 + 2} \right) \right\}\]

\[ = \left\{ \left( - \frac{4^2}{{27}_3} \times \frac{9}{2} \right)\left( x^{1 + 2} y^{1 + 1} z^{1 + 1} \right) \right\} - \left\{ \left( - \frac{4^1}{{27}_9} \times \frac{3}{4} \right)\left( x^{1 + 1} y^{1 + 1} z^{1 + 2} \right) \right\}\]

\[ = - \frac{2}{3} x^3 y^2 z^2 + \frac{1}{9} x^2 y^2 z^3\]

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