 The solution to a cubic is a good example of a key process in mathematics, change things up. Let's consider, to solve a quadratic equation like x squared plus 10x equals 39, we add a constant, the square of half the coefficient of x, and this completes the square. So the natural thing to try is to add a constant to complete the cube, but we can't. And the problem is that a cube like x cubed plus 6x equals 20 requires the addition of both x squared and constant terms to become a perfect cube. So now what? Well, let's take a different perspective. The reason we're adding something to complete the square is that the quantity we have is an incomplete square. And so we asked, what do we add to the incomplete square to make a complete square? Well, let's change things up. What if we asked, what could we subtract from the complete square to make an incomplete square? In other words, suppose our variable expression is the difference of two squares. So suppose we want to solve x squared plus 10x equals 39. The variable part x squared plus 10x can be viewed as the difference of squares, u squared minus v squared. And so we see that this x, the unknown side, is actually u minus v. But if the big square is u squared, then u squared minus v squared, that gets us our x squared plus 10x. But if we then subtract x squared, we leave 10x. What's left are these two wings, which are 2vx. And so this means that 10x has to be 2vx. And so we know that u squared minus v squared equals 39, the constant, 2v equals 10, the coefficient of x, and x itself is u minus v. And this allows us to solve the equation. So since 2v is equal to 10, we know that v is equal to 5. And since u squared minus v squared equals 39, and we know what v is, we can find u. And since x is u minus v, we can find x. Can we do this with x cubed plus 6x equals 20? And our goal is we'd like x cubed plus 6x to be u cubed minus v cubed, the difference of two cubes. So we have lengths x, u, and v. And if we explode our cube, we see that it's a cube of x plus three slabs with dimension x, u, and v. And so we need 6x to equal 3uv, and this gives us uv equals 2. And notice that we have u cubed minus v cubed equals 20, a difference. And a product, well not quite the product that we need, we have the product uv equals 2. But if we cube it, we get u cubed v cubed equals 8. So we have a difference and a product, and this is a mesopotamian canal problem. At least it can be solved as one. So we might start out this way. The product of two numbers is 8, and their difference is 20. So the difference is 20. So half the difference is 10, squared is 100. If we add the product, we get the square of the sum is equal to 108. We can then take the square root, and so we can find that half the sum is the square root of 108. And the thing to realize here is that the thing we're adding or subtracting are actually the cubes of u and v. And so our work gives us half the difference and half the sum. And since we know half the sum and half the difference, if we add them, we'll get the first. And so we can take the cube root to find u. We can also subtract to find the other one. And finally, since x equals u minus v, we have our solution.