# How do you factor 0.064a^3+8y^9?

Apr 12, 2015

When asked to factor $0.064 {a}^{3} + 8 {y}^{9}$, the first things I notice are the addition and the ${a}^{3}$ -- a cube.
Then "Oh, look, $8$ is a cube and, so is ${y}^{9}$".

(${2}^{3} = 8$ and ${\left({y}^{3}\right)}^{3} = {y}^{9}$)

So, I'm thinking maybe it's a sum of two cubes.

But what about that $0.064$?

Well, 3 right of the decimal is 1000ths and that's dividing by 1,000 which is ${10}^{3}$, that is: $\frac{\textcolor{w h i t e}{64}}{1000} = \frac{\textcolor{w h i t e}{64}}{10} ^ 3$

OK, so $0.064 = \frac{64}{100} = \frac{64}{10} ^ 3$

So all I'm left wondering about is the $64$

I cross my fingers and start trying numbers.
${2}^{3}$ is only $8$,
${3}^{3}$ is an odd number, it can't be $64$
${4}^{3} = 4 \cdot 16 = 64$ and there it is:

$0.064 {a}^{3} + 8 {y}^{9} = \textcolor{red}{{\left(\frac{4}{10} a\right)}^{3}} + {\left(2 {y}^{3}\right)}^{3}$

Use the rule I've memorized:

$\textcolor{red}{{u}^{3}} + {v}^{3} = \left(\textcolor{red}{u} + v\right) \left(\textcolor{red}{{u}^{2}} - \textcolor{red}{u} v + {v}^{2}\right)$ to get:

$\textcolor{red}{{\left(\frac{4}{10} a\right)}^{3}} + {\left(2 {y}^{3}\right)}^{3} = \left[\textcolor{red}{\left(\frac{4}{10} a\right)} + \left(2 {y}^{3}\right)\right] \left[\textcolor{red}{{\left(\frac{4}{10} a\right)}^{2}} + \textcolor{red}{\left(\frac{4}{10} a\right)} \left(2 {y}^{3}\right) + {\left(2 {y}^{3}\right)}^{2}\right]$

$= \left[\left(\frac{4}{10} a\right) + \left(2 {y}^{3}\right)\right] \left[\left(\frac{16}{100} {a}^{2}\right) + \left(\frac{4}{10} a\right) \left(2 {y}^{3}\right) + \left(4 {y}^{6}\right)\right]$

$= \left[0.4 a + 2 {y}^{3}\right] \left[0.16 {a}^{2} + 0.4 a \cdot 2 {y}^{3} + 4 {y}^{6}\right]$

$= \left[0.4 a + 2 {y}^{3}\right] \left[0.16 {a}^{2} + 0.8 a {y}^{3} + 4 {y}^{6}\right]$.