# Math Help - Cubed Root Question

1. ## Cubed Root Question

In the following factoring question

$64x^3 - 27y^6$

(a - b) (a^2 + ab + b^2) is used

However, I am stuck right off the bat. The cube root of $64x^3$ is 4x, but does one do the cubed root of $27y^6$. Its obviously a difference of perfect cubes because x^3 and y^6 are both multiples of 3. The book tells me the cubed root of $27^6$ is $3y^2$, but I do not know specifically how the $y^6$turns into $y^2$ (obviously the cube root of 27 is 3).

Could some one please clarify and show me?

2. Hello,

Remember the rules of the powers :

$x^{yz}=(x^y)^z=(x^z)^y$

So here :
$y^6=y^{2 \cdot 3}=(y^2)^3$

3. For a better understanding lets take this same term but out of the cubed root application.

If I was to factor
$y^6$ by itself, how would I know the correct order to factor it? I could do $(y^{3*2}) = (y^3)^2$ instead of the correct order $y^{2*3} = (y^2)^3$

Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

Thanks

4. Originally Posted by allyourbass2212
For a better understanding lets take this same term but out of the cubed root application.

If I was to factor
$y^6$ by itself, how would I know the correct order to factor it? I could do $(y^{3*2}) = (y^3)^2$ instead of the correct order $y^{2*3} = (y^2)^3$

Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

Thanks
Both are correct

So you can take $(y^2)^3$ or $(y^3)^2$, according to what you need

5. Originally Posted by allyourbass2212
For a better understanding lets take this same term but out of the cubed root application.

If I was to factor
$y^6$ by itself, how would I know the correct order to factor it? I could do $(y^{3*2}) = (y^3)^2$ instead of the correct order $y^{2*3} = (y^2)^3$

Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

Thanks
The idea here is to have a base raised to the 3rd power. That's what your rule requires: the difference of two cubes.

$(a^3-b^3)$

If you used $(y^3)^2$ , you wouldn't have a base cubed. You would have a base squared.