# How do you factor 5x^3 + 40y^6?

##### 1 Answer
May 18, 2015

You can use a formula for a sum of cubes:
${a}^{3} + {b}^{3} = \left(a + b\right) \cdot \left({a}^{2} - a b + {b}^{2}\right)$

Notice that $40 {y}^{6} = 5 \cdot 8 \cdot {y}^{6} = 5 \cdot {2}^{3} \cdot {\left({y}^{2}\right)}^{3} = 5 \cdot {\left(2 {y}^{2}\right)}^{3}$

Using the above formula for $a = x$ and $b = 2 {y}^{2}$ and factoring out constant $5$, we get:
5x^3+40y^6 =5x^3+5(2y^2)^3=5(x+2y^2)(x^2-2xy^2+4y^4))