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January 3rd, 2019, 06:47 PM   #1
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Solving roots equation neatly

I am to find all real solutions for this problem.

$2\sqrt[3]{x^2}-\sqrt[3]{x}=1$

According to my textbook, the solutions are {-$\frac{1}{8}$, 1}. However, I keep getting {-$\frac{1}{2}$, 1} when I try to solve it more practically. I went on Symbolab to see how they do it and I learned if I use the exponent property $a^{n}=(\sqrt[m]{a})^{n\times m}$, I get the correct solution set. My issue with this, though, is that I'm not going to be able to remember that exponent property well and the work after that gets too tedious for this kind of problem.

Here's one of my attempts of solving this problem to give you an idea of what I've been doing.
$$2\sqrt[3]{x^2}-\sqrt[3]{x}=1$$
$$=2x^{\frac{2}{3}}-x^{\frac{1}{3}}=1$$
$$=(2x^{\frac{2}{3}}-x^{\frac{1}{3}})^3=1^3$$
$$=2x^2-x-1=0$$
$$=x(2x+1)-1(2x+1)=0$$
$$=(x-1)(2x+1)=0$$
$$x={-\frac{1}{2}, 1}$$

Of course that's a flawed solution, but can anyone help me find a more elegant approach to solving this problem? Please use TeX commands
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January 3rd, 2019, 06:53 PM   #2
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$u = \sqrt[3]{x}$

$2u^2 - u - 1 = 0$

$(2u+1)(u-1) = 0$

$u = -\dfrac 1 2,~u=1$

$x = u^3$

$x = -\dfrac 1 8,~x=1$
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January 3rd, 2019, 06:54 PM   #3
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$$(a+b)^3 \ne a^3 + b^3$$
Specifically, $$(2x^\tfrac23 - x^\tfrac13)^3 \ne 2x^2-x$$
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January 3rd, 2019, 07:02 PM   #4
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Wow. I can't believe I made that mistake. I'll try to actually expand it and see where it goes from there.
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January 3rd, 2019, 07:15 PM   #5
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Letting $x = y^3\!$, so that $y = \sqrt[3]{x}$, gives $2y^2 - y - 1 = 0$. Hence $y = -1/2$ or $1$, and so $x = -1/8$ or $1$.
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