Solve 16a^4x^3+54ay^3

$$16 a ^ { 4 } x ^ { 3 } + 54 a y ^ { 3 }$$

Factor

$$2a\left(2ax+3y\right)\left(4a^{2}x^{2}-6axy+9y^{2}\right)$$

Solution Steps

Factor out 2.

$$2\left(8a^{4}x^{3}+27ay^{3}\right)$$

Consider 8a^{4}x^{3}+27ay^{3}. Factor out a.

$$a\left(8a^{3}x^{3}+27y^{3}\right)$$

Consider 8a^{3}x^{3}+27y^{3}. Rewrite 8a^{3}x^{3}+27y^{3} as \left(2ax\right)^{3}+\left(3y\right)^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

$$\left(2ax+3y\right)\left(4a^{2}x^{2}-6axy+9y^{2}\right)$$

Rewrite the complete factored expression.

$$2a\left(2ax+3y\right)\left(4a^{2}x^{2}-6axy+9y^{2}\right)$$

Evaluate

$$54ay^{3}+16a\left(ax\right)^{3}$$