# Difference of cubes?

### Difference of cubes?

The difference of cubes states that a^3-b^3 = (a-b)(a^2+ab+b^2)

So I have the problem to factor, which is:

y^6-13y^3+40

I factored and got this:

(y^3-8)(y^3-5)

That first term is the difference of cubes so the formula should work, but I used it and got:

(y-2)(y^2+2y+4)(y^3-5) BUT.....

my answer key says it should be (y-2)(y^4+2y+4)(y^3-5)

I don't understand why it's y^4 here. It's probably simple, I know, but I've been doing this stuff all day and I think my brain is fried. I'm in my 30's trying to reteach myself this stuff after a long time of ignoring it. Any help would be appreciated. Thanks.
Guest

### Re: difference of cubes?

Good morning!

See:
$$y^6-13y^3+40$$ is sixth degree, right?
$$\left(y-2\right)\cdot\left(y^2+2y+4\right)\cdot\left(y^3-5\right)$$

If you multiply $$y\cdot y^2\cdot y^3=y^6$$, right?

What happens if you multiply by a $$y^4$$, instead of $$y^2$$?

You got a wrong degree answer (8)!

I hope I have helped!

Baltuilhe

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Joined: Fri Dec 14, 2018 3:55 pm
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### Re: Difference of cubes?

One quick check is to add the highest powers. In (y- 2)(y^4+ 2y+ 4)(y^3- 5), the "leading power" is 1+ 4+ 3= 8. But the original polynomial had leading power 6, not 8. Yes, it's simple- a simple typo in the answer!
Guest