How do you factor the expression 16x^3 + 54y^6?

Dec 20, 2015

Factor out a $2$ and apply the sum of cubes formula to find that

$16 {x}^{3} + 54 {y}^{6} = 2 \left(2 x + 3 {y}^{2}\right) \left(4 {x}^{2} - 6 x {y}^{2} + 9 {y}^{4}\right)$

Explanation:

The sum of cubes formula states that

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

(verify this by expanding the right hand side)

With this available, we have

$16 {x}^{3} + 54 {y}^{6} = 2 \left(8 {x}^{3} + 27 {y}^{6}\right)$

$= 2 \left({2}^{3} {x}^{3} + {3}^{3} {\left({y}^{2}\right)}^{3}\right)$

$= 2 \left({\left(2 x\right)}^{3} + {\left(3 {y}^{2}\right)}^{3}\right)$

(applying the sum of cubes formula)

$= 2 \left(2 x + 3 {y}^{2}\right) \left({\left(2 x\right)}^{2} - \left(2 x\right) \left(3 {y}^{2}\right) + {\left(3 {y}^{2}\right)}^{2}\right)$

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