Solve 3b^4c-18b^3c^2+27b^2{c}^3

$$3 { b }^{ 4 } c-18 { b }^{ 3 } { c }^{ 2 } +27 { b }^{ 2 } { c }^{ 3 }$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(b^{4}c-6b^{3}c^{2}+9b^{2}c^{3}\right)$$

Consider b^{4}c-6b^{3}c^{2}+9b^{2}c^{3}. Factor out b^{2}c.

$$b^{2}c\left(b^{2}-6bc+9c^{2}\right)$$

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

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

Rewrite the complete factored expression.

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

Evaluate

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