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 January 3rd, 2019, 06:47 PM #1 Newbie   Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0 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
 January 3rd, 2019, 06:53 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,575 Thanks: 1423 $u = \sqrt[3]{x}$ $2u^2 - u - 1 = 0$ $(2u+1)(u-1) = 0$ $u = -\dfrac 1 2,~u=1$ $x = u^3$ $x = -\dfrac 1 8,~x=1$ Thanks from sigma
 January 3rd, 2019, 06:54 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra $$(a+b)^3 \ne a^3 + b^3$$ Specifically, $$(2x^\tfrac23 - x^\tfrac13)^3 \ne 2x^2-x$$ Thanks from sigma
 January 3rd, 2019, 07:02 PM #4 Newbie   Joined: Oct 2018 From: arizona Posts: 6 Thanks: 0 Wow. I can't believe I made that mistake. I'll try to actually expand it and see where it goes from there.
 January 3rd, 2019, 07:15 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 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 $1$. Thanks from greg1313

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