Solve 3a^3-48ab^4=

$$3 { a }^{ 3 } -48a { b }^{ 4 } =$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(a^{3}-16ab^{4}\right)$$

Consider a^{3}-16ab^{4}. Factor out a.

$$a\left(a^{2}-16b^{4}\right)$$

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

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

Rewrite the complete factored expression.

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

Evaluate

$$3a\left(a^{2}-16b^{4}\right)$$