Solve 16a^3bc-49ab^3c

$$16 a ^ { 3 } b c - 49 a b ^ { 3 } c$$

Factor

$$abc\left(4a-7b\right)\left(4a+7b\right)$$

Solution Steps

Factor out abc.

$$abc\left(16a^{2}-49b^{2}\right)$$

Consider 16a^{2}-49b^{2}. Rewrite 16a^{2}-49b^{2} as \left(4a\right)^{2}-\left(7b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(4a-7b\right)\left(4a+7b\right)$$

Rewrite the complete factored expression.

$$abc\left(4a-7b\right)\left(4a+7b\right)$$

Evaluate

$$abc\left(16a^{2}-49b^{2}\right)$$