Question

Multiply the following:

$\left(-2a^{2}b\right)\left(3a{b}^2+2b{c}^2-\frac{4}{9}{c}^{2}a+2a^2{b}^2-3b^2{c}^2{a}^2\right)$

Answer 1

Simplify given expression$$\begin{gathered} \left(-2a^{2}b\right)\left(3a{b}^2+2b{c}^2-\frac{4}{9}{c}^{2}a+2a^2{b}^2-3b^2{c}^2{a}^2\right)=\left(-2a^{2}b\right)\left(3a{b}^2\right)+\left(-2a^{2}b\right)\left(2b{c}^2\right)+\left(-2a^{2}b\right)\left(-\frac{4}{9}{c}^{2}a\right)+\left(-2a^{2}b\right)\left(2a^2{b}^2\right)+\left(-2a^{2}b\right)\left(-3b^2{c}^2{a}^2\right) \\ =\left(-2\times 3\right)\left(a^{2}b\times a{b}^2\right)+\left(-2\times 2\right)\left(a^{2}b\times b{c}^2\right)+\left(-2\times -\frac{4}{9}\right)\left(a^{2}b\times c^{2}a\right)+\left(-2\times 2\right)\left(a^{2}b\times a^2{b}^2\right)+\left(-2\times -3\right)\left(a^{2}b\times b^2{c}^2{a}^2\right) \\ =-6a^3{b}^3-4a^2{b}^2{c}^2+\frac{8}{9}{a}^{3}b{c}^2-4a^4{b}^3+6a^4{b}^3{c}^2 \end{gathered}$$

Hence, the product of algebraic expression is $-6a^3{b}^3-4a^2{b}^2{c}^2+\frac{8}{9}{a}^{3}b{c}^2-4a^4{b}^3+6a^4{b}^3{c}^2$.