Solve 3a^4-48b^4

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

Factor

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

Solution Steps

Factor out 3.

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

Consider a^{4}-16b^{4}. Rewrite a^{4}-16b^{4} as \left(a^{2}\right)^{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).

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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