Is there some way to express this as a quadratic equation? Create a new variable, *u*, and set it equal to *x*^{1/3}. *u*^{2} + *u* + 1 = 0
Hey hey hey, this new equation is quadratic. It isn't factorable, though, so we use the quadratic formula to find the roots. and These are the roots for the new, *u*-based equation. Since we already said that *u* = *x*^{1/3}, solving for *x* gives us *x* = *u*^{3}. and We were right; this is a super nasty problem. Let's go ahead and check our results, though. Luckily, most of the exponents cancel each other out. One down. And the other. That means our answers are and . It's really satisfying to get a long problem like this right after all this work. It gives us a warm, fuzzy feeling all over. |