I'm going around in circles on this one...
Solve the equation $$x^{2/3}-6x^{1/3} + 8 = 0$$
According to the book, the answer is 8, 64, but that makes no sense to me either...
Thanks
Gary
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$\begingroup$Let $y=x^{1/3}$, then the equation is $$y^2-6y+8=0$$ which factors nicely. Can you take it from here?
$\endgroup$ 1 $\begingroup$This is just a quadratic in disguise. Don't you see that $x^{2/3}$ can be actually rewritten like this: $\left(x^{\frac{1}{3}}\right)^2$? There are typically two strategies that are used to approach solving this type of equation. You either assign your original expression under the square to a new variable and solve the equation with respect to that new variable and when done substitute back what you've got and finally solve it using the original expression. Or if you anything like me, you just don't care and treat $x^{1/3}$ as a single quantity.
$$ x^\frac{2}{3}-6x^\frac{1}{3} + 8 = 0\implies\\ \left(x^{\frac{1}{3}}\right)^2-6x^\frac{1}{3} + 8 = 0\implies\\ \left(x^{\frac{1}{3}}\right)^2-2\cdot 3\cdot x^\frac{1}{3} +3^2 -3^2 + 8 = 0\implies\\ \left(x^{\frac{1}{3}} - 3\right)^2=9-8\implies\\ \left(x^{\frac{1}{3}} - 3\right)^2=1\implies\\ x^{\frac{1}{3}} - 3=\pm 1\implies\\ x^{\frac{1}{3}} =3\pm 1\implies\\ x^{\frac{1}{3}} =3+ 1 \;\;\;\;\;or \;\;\;\;\;x^{\frac{1}{3}} =3- 1\implies\\ x^{\frac{1}{3}} =4 \;\;\;\;\;or \;\;\;\;\;x^{\frac{1}{3}} =2\implies\\ x =4^3 \;\;\;\;\;or\;\;\;\;\; x =2^3\implies\\ x =64 \;\;\;\;\;or \;\;\;\;\;x =8 $$
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