Factoring Madness: Intersecting Curves

[zeion]

Homework Statement

I need to find where these curves intersect:

x+y = 2y^2
y = x^3

Homework Equations

The Attempt at a Solution

So I try to write both in terms of y, I get

2y^2 - y = y^(1/3)

I try to factor

2y^2 - y - y^(1/3) = 0
y(2y - 1 - y^(-2/3)) = 0 ??

What do I do next?

 

[Mark44]

Homework Statement

I need to find where these curves intersect:

x+y = 2y^2
y = x^3

Homework Equations

The Attempt at a Solution

So I try to write both in terms of y, I get

2y^2 - y = y^(1/3)

You're on the right track up to the step above, but that's not an equation that's easy to factor with that cube root.

Instead, cube both sides of the equation. The right side is easy; when cubed it becomes just y.
The left side requires some care, because you're multiplying (2y² - y)(2y² - y)(2y² - y). When you're done, the highest degree term will be 8y⁶. This might turn out to be a messy problem, since you'll be needing to factor a 6th degree polynomial to find the intersection point(s) - I believe there are two.

I try to factor

2y^2 - y - y^(1/3) = 0
y(2y - 1 - y^(-2/3)) = 0 ??

What do I do next?

 

[zeion]

Okay so after I cubed both sides I get:

8y⁶ - 12y⁵ - 2y⁴ - y³ - y =0

Then I factor out a y:

y(8y⁵ - 12y⁴ - 2y³ - y² - 1) = 0

How do I factor now..? Was there some way to check easier with ration roots theorem?

 

[Mentallic]

Yes the other root is rational so use the theorem to find it.

It would have saved you a lot of hassle cubing and even makes finding the other root besides y=0 much easier if you substituted $y=x^3$ into $x+y=2y^2$. You will quickly and easily get $x+x^3=2x^6$.