## College Algebra (6th Edition)

$x = 27$ or $x = -8$
We can re-write the equation in the following manner using the rules of exponents: $$x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0$$ $$(x^{\frac{1}{3}})^{2} - (x^{\frac{1}{3}}) - 6 = 0$$ $$b^{2} - b - 6 = 0$$ where $b = x^{\frac{1}{3}}$. Since 2 factors of 6 that, when substracted give 1 are 2 and 3, we can factorize our new equation as so: $$0 = (b-3)(b+2)$$ and solve for $b$: $b - 3 = 0$ or $b+2 = 0$ $b = 0+3$ or $b=0-2$ $b = 3$ or $b=-2$ Since we had established that $b = x^{\frac{1}{3}}$, we can now solve for $x$: $x^{\frac{1}{3}} = 3$ or $x^{\frac{1}{3}} = -2$ $(x^{\frac{1}{3}})^{3} = 3^{3}$ or $(x^{\frac{1}{3}})^{3} = (-2)^{3}$ $x = 27$ or $x = -8$ where, if we substitute the values into the original equation, we conclude that both satisfy the original conditions.