Solving roots equation neatly

Oct 2018
7
0
arizona
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
 

romsek

Math Team
Sep 2015
2,959
1,673
USA
$u = \sqrt[3]{x}$

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

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

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

$x = u^3$

$x = -\dfrac 1 8,~x=1$
 
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v8archie

Math Team
Dec 2013
7,712
2,682
Colombia
$$(a+b)^3 \ne a^3 + b^3$$
Specifically, $$(2x^\tfrac23 - x^\tfrac13)^3 \ne 2x^2-x$$
 
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Oct 2018
7
0
arizona
Wow. I can't believe I made that mistake. I'll try to actually expand it and see where it goes from there.
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
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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