Solve 3a^5b-18a^3b^3+27ab^5=

$$3 a ^ { 5 } b - 18 a ^ { 3 } b ^ { 3 } + 27 a b ^ { 5 } =$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(a^{5}b-6a^{3}b^{3}+9ab^{5}\right)$$

Consider a^{5}b-6a^{3}b^{3}+9ab^{5}. Factor out ab.

$$ab\left(a^{4}-6a^{2}b^{2}+9b^{4}\right)$$

Consider a^{4}-6a^{2}b^{2}+9b^{4}. Use the perfect square formula, p^{2}-2pq+q^{2}=\left(p-q\right)^{2}, where p=a^{2} and q=3b^{2}.

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

Rewrite the complete factored expression.

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

Evaluate

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