Solve 2a^3-3a^2b-32ab^2+48b^3

$$2 a ^ { 3 } - 3 a ^ { 2 } b - 32 a b ^ { 2 } + 48 b ^ { 3 }$$

Factor

$$\left(a-4b\right)\left(2a-3b\right)\left(a+4b\right)$$

Solution Steps

Do the grouping 2a^{3}-3a^{2}b-32ab^{2}+48b^{3}=\left(2a^{3}-3a^{2}b\right)+\left(-32ab^{2}+48b^{3}\right), and factor out a^{2} in the first and -16b^{2} in the second group.

$$a^{2}\left(2a-3b\right)-16b^{2}\left(2a-3b\right)$$

Factor out common term 2a-3b by using distributive property.

$$\left(2a-3b\right)\left(a^{2}-16b^{2}\right)$$

Consider a^{2}-16b^{2}. Rewrite a^{2}-16b^{2} as a^{2}-\left(4b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-4b\right)\left(a+4b\right)$$

Rewrite the complete factored expression.

$$\left(a-4b\right)\left(a+4b\right)\left(2a-3b\right)$$

Evaluate

$$\left(2a-3b\right)\left(a^{2}-16b^{2}\right)$$