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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?

Joe
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munk
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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 02 '12 at 04:35
  • 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. – munk Jun 02 '12 at 04:46
  • 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 02 '12 at 04:47
  • How does knowing a root of a cubic reduce the problem to a quadratic? Can you clarify that a little? – munk Jun 02 '12 at 04:53
  • I was referring to the use of synthetic division. Consider reading up on http://www.mash.dept.shef.ac.uk/Resources/web-cubicequations-john.pdf Possibly I should have not written the comment about reducing a cubic to a quadratic since it does not directly apply to this question (but the strategy for inspection does, that is why I did.) Sorry if I confused you at all. – Joe Jun 02 '12 at 04:58
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    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! – munk Jun 02 '12 at 05:00
  • Your equation reminds you of a quadratic because it's an equation of "quadratic type." This means a suitable substitution transforms it into a quadratic, in this case $y= \sqrt3$. This is what user9413 suggested in the answer you accepted. – Chris Leary May 23 '13 at 22:46

1 Answers1

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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.