Solve 81p^3q^2-24q^2

$$81 p ^ { 3 } q ^ { 2 } - 24 q ^ { 2 }$$

Factor

$$3\left(3p-2\right)q^{2}\left(9p^{2}+6p+4\right)$$

Solution Steps

Factor out 3.

$$3\left(27p^{3}q^{2}-8q^{2}\right)$$

Consider 27p^{3}q^{2}-8q^{2}. Factor out q^{2}.

$$q^{2}\left(27p^{3}-8\right)$$

Consider 27p^{3}-8. Rewrite 27p^{3}-8 as \left(3p\right)^{3}-2^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(3p-2\right)\left(9p^{2}+6p+4\right)$$

Rewrite the complete factored expression. Polynomial 9p^{2}+6p+4 is not factored since it does not have any rational roots.

$$3q^{2}\left(3p-2\right)\left(9p^{2}+6p+4\right)$$

Evaluate

$$3q^{2}\left(27p^{3}-8\right)$$