Question

By letting y = x^1/3, or otherwise, find the values of x for which x^1/3 -2x^-1/3 = 1

Answer 1

In this case, the problem is asking us to solve the equation

$x^{\frac{1}{3}} - \dfrac{2}{x^{\frac{1}{3}}} = 1$

by substituting $y = x^{\frac{1}{3}}$.

First, let's substitute the variable into the original equation:

$y - \dfrac{2}{y} = 1$

Now, let's solve:

$y - \dfrac{2}{y} = 1$

• Set up

$y\Bigg(y - \dfrac{2}{y}\Bigg) = y(1) \Rightarrow y^2 - 2 = y$

• Multiply the entire equation by $y$ to remove the fraction over $y$

$y^2 - y - 2 = 0$

• Subtract $y$ from both sides of the equation for easier factoring

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

• Factor

$y - 2 = 0 \,\, \textrm{and} \,\, y + 1 = 0$

• Solve the factored equation using the Zero Product Property

$y = 2, -1$

• Solve each equation

Now, we aren't done yet. We have to find our answer in terms of the original variable, $x$. To do this, set each value for $y$ into the substitution equation we found earlier, $y = x^{\frac{1}{3}}$:

$2 = x^{\frac{1}{3}}$

$2^3 = x$

$x = 8$

$-1 = x^{\frac{1}{3}}$

$(-1)^3 = x$

$x = -1$

The solutions for the equation are x = -1, 8.