### Video Transcript

Find the solution set of the cube root of 𝑥 to the fifth power minus 244 equals negative 243 over the cube root of 𝑥 to the fifth power.

To find the solutions of this equation, we need to find the values of 𝑥 that make it true. But at the moment, this looks like quite a nasty equation. So we’re going to begin by dealing with this denominator. We’ll begin by multiplying both sides of our equation by the cube root of 𝑥 to the fifth power. The cube root of 𝑥 to the fifth power times itself can be written as the cube root of 𝑥 to the fifth power squared. Now, we could write this as a single exponent of 𝑥, but actually it’s much easier to keep it like this for now.

We multiply negative 244 by the cube root of 𝑥 to the fifth power. And when we multiply the term on the right-hand side by this value, we’re simply left with negative 243. Next, we’re going to add 243 to both sides of our equation. So we have the cube root of 𝑥 to the fifth power squared minus 244 times the cube root of 𝑥 to the fifth power plus 243 equals zero. And you might notice this looks a little like a quadratic equation. To make this a little more obvious, we’re going to perform a substitution. We’re going to let 𝑦 be equal to the cube root of 𝑥 to the fifth power. And so our equation becomes 𝑦 squared minus 244𝑦 plus 243 equals zero. And we know that to solve this quadratic equation, one of the techniques we can use is to factor the expression on the left-hand side.

The first term in each binomial must be 𝑦, since 𝑦 times 𝑦 is 𝑦 squared. Then, we’re looking for two numbers whose product is positive 243 and whose sum is negative 244. Well, these values are negative 243 and negative one. So our equation becomes 𝑦 minus 243 times 𝑦 minus one equals zero. For the product of these two binomials to be equal to zero, either 𝑦 minus 243 must be equal to zero or 𝑦 minus one must be equal to zero. And so we solve this first equation by adding 243 to both sides, so 𝑦 is equal to 243. And we solve the second equation by adding one to both sides, so 𝑦 is equal to one.

But, of course, we said we’re looking to find the values of 𝑥 which make this equation true. So we go back to our earlier substitution, 𝑦 is equal to the cube root of 𝑥 to the fifth power. And we can say that either the cube root of 𝑥 to the fifth power must be equal to 243 or one. Well, the second equation is fairly straightforward to solve. We know that 𝑥 must be equal to one. But how do we solve the first equation. The cube root of 𝑥 to the fifth power equals 243. Well, let’s begin by taking the fifth root of both sides of our equation. The fifth root of 243 is three. So we find that the cube root of 𝑥 must be three. We’re then going to cube both sides. And we find that 𝑥 is equal to 27. And so there are two values in our solution set. They are one and 27.