Difference of cubes?

[Guest]

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.

[Baltuilhe]

Good morning!

Your answer is correct!

See:
[tex]y^6-13y^3+40[/tex] is sixth degree, right?
Your answer:
[tex]\left(y-2\right)\cdot\left(y^2+2y+4\right)\cdot\left(y^3-5\right)[/tex]

If you multiply [tex]y\cdot y^2\cdot y^3=y^6[/tex], right?

What happens if you multiply by a [tex]y^4[/tex], instead of [tex]y^2[/tex]?

You got a wrong degree answer (8)!

So, ask to your teacher to review the answer, because it's totally wrong

I hope I have helped!

[Guest]

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!

[GeradHum]

Sure! Multiply both sides by (x+2)(2x+1):
1 = a(2x + 1) + b(x + 2)
Expand:
1 = 2ax + a + bx + 2b
Group x terms:
1 = x(2a + b) + (a + 2b)
Set coefficients equal:
For x: 2a + b = 0
Constant: a + 2b = 1
Solve system:
From 2a + b = 0 → b = -2a
Plug into second: a + 2(-2a) = 1 → a - 4a = 1 → -3a = 1 → a = -1/3
Then b = -2(-1/3) = 2/3
So: a = -1/3, b = 2/3

[Poeth1952]

Don’t worry, it’s totally normal to get a bit confused. Let’s break it down step by step. Your expression is:

y^6 - 13y^3 + 40


First, notice it’s like a cubic in disguise because you can set u = y^3. Then it becomes simpler:

u^2 - 13u + 40


Factor it like a normal quadratic: find two numbers that multiply to 40 and add to -13 → that’s -5 and -8. So we get:

(u - 8)(u - 5)


Substitute back u = y^3 and we have:

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


The first term (y^3 - 8) is a difference of cubes (a^3 - b^3 = (a-b)(a^2 + ab + b^2)), with a = y and b = 2, so it factors as:

y^3 - 8 = (y - 2)(y^2 + 2y + 4)


Then the whole thing factors like this:

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


About that “y^4” in the answer key — that’s just a mistake. The difference of cubes always gives a quadratic factor (y^2 + 2y + 4), not a quartic.

So your answer is correct, and don’t stress — it’s totally normal to get tired after studying this stuff all day.