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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?

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    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 $y= \sqrt[3](x)$. This is what user9413 suggested in the answer you accepted.2013-05-23

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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.