Solve 3a^2b^3-27a^4b

$$3 a ^ { 2 } b ^ { 3 } - 27 a ^ { 4 } b$$

Factor

$$3b\left(b-3a\right)\left(3a+b\right)a^{2}$$

Solution Steps

Factor out 3.

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

Consider a^{2}b^{3}-9a^{4}b. Factor out a^{2}b.

$$a^{2}b\left(b^{2}-9a^{2}\right)$$

Consider b^{2}-9a^{2}. Rewrite b^{2}-9a^{2} as b^{2}-\left(3a\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(b-3a\right)\left(b+3a\right)$$

Reorder the terms.

$$\left(-3a+b\right)\left(3a+b\right)$$

Rewrite the complete factored expression.

$$3a^{2}b\left(-3a+b\right)\left(3a+b\right)$$

Evaluate

$$3ba^{2}\left(b^{2}-9a^{2}\right)$$