# Solving $\sqrt[3]{x^2} + \sqrt[3]{x} = 2$

My sister asked me for some help on her algebra homework the other day, and I was stumped by her question. The problem is to find the root of $\sqrt[3]{x^2} + \sqrt[3]{x} = 2$.

The internet tells me that x is 1, but I can't seem to figure out why.

I've tried to manipulate it a couple of ways and I always end up with something more complex than when I started like $8 - 12x^{1/3} + 6x^{2/3} - x - x^2 = 0$.

The simplest form I've found is $x^{2/3} + x^{1/3} - 2 = 0$ which reminds me of the quadratic formula a little, but the exponents are rational, not quite what I need.

Any advice on how to proceed?

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The first thing I would do with equations like this is to look for small integer solutions. From my view, it easy to see that $x=1$ is a solution. Beware there may not always be small integer solutions. –  Joe Jun 2 '12 at 4:35
I am more interested in a general solution, not something that is trial and error. It's a good method, but not quite what I'm looking for. –  munkhd Jun 2 '12 at 4:46
I know. I felt the answer below was fine for that, so I wanted to let you know that simply by inspection helps determine a root quickly, say for deducing a cubic to a quadratic using the fact that you know one of the roots (for future reference). –  Joe Jun 2 '12 at 4:47
How does knowing a root of a cubic reduce the problem to a quadratic? Can you clarify that a little? –  munkhd Jun 2 '12 at 4:53
Maybe it's not related, but it's something new to me that I might not have discovered on my own, so I appreciate where it lead! –  munkhd Jun 2 '12 at 5:00

Put $x= y^{3}$ then your equation reduces to $$y^{2}+y -2 = (y+2)(y-1)=0$$ So from here you get $y=-2$ or $y=1$.
• So if $y=1$, then you get $x=1$.
• And if $y=-2$, you get $x=-8$, which also satisfies the equation.