Sum and difference of cubes, and the three-term cubic identity

0:00Two products, written down without an answer. x minus y, times x squared plus x y plus y squared. And x plus y, times x squared minus x y plus y squared. Six terms each. No obvious reason to care. But one of them collapses to two terms, and the other collapses the same way with every sign reversed. That is not luck. Somebody built the second bracket so it would happen.

0:33Multiply the first one out. Every term on the left against every term on the right. x gives x cubed, x squared y, x y squared. Minus y gives minus x squared y, minus x y squared, minus y cubed. Six products. Now look at the middle four. Plus x squared y, and minus x squared y. Gone. Plus x y squared, and minus x y squared. Gone. Four dead, in two pairs. What is left is x cubed minus y cubed, and nothing else at all.

1:14So why exactly four? Look at the second bracket. The power of x runs two, one, nought. The power of y runs nought, one, two. It steps by one. Multiplying by x pushes every term up a step in x. Minus y pushes every term up a step in y, and flips the sign. Same size step. So the second thing x makes lands exactly on the first thing minus y makes. Same shape, opposite sign.

1:47The middles pair off, one from each side. And the two ends have nobody to meet. x times x squared is off the top. Minus y times y squared is off the bottom. Two survivors, because only two of the six were ever unpaired. That is the design. Now the second product, without multiplying it out. Flip the sign outside and flip the middle inside, and the same pairing runs with every sign reversed.

2:20x plus y, times x squared minus x y plus y squared, is x cubed plus y cubed. And the pairing is forced. I tried all four ways of choosing those two signs. Two collapse to two terms. The other two give four terms and no factorisation at all. The two that work are the ones where the signs are opposite. Minus outside wants plus in the middle. Get it backwards and nothing cancels. That is the standard mistake, and it is not a memory failure.

2:57A warning, because these two get confused more than anything else here. x cubed minus y cubed is not x minus y, all cubed. That one is x cubed minus three x squared y plus three x y squared minus y cubed. Four terms. I checked them against each other on forty nine pairs of numbers. They agree on nineteen — every one with a letter at nought, or the two equal.

3:27They agree only where almost anything agrees. And x squared plus x y plus y squared is not a perfect square either. A square would have two x y in the middle. This has one, and it does not break up further. x minus y divides x squared minus y squared, and now x cubed minus y cubed. What about the fourth power? Split it as a difference of squares, and split the first half again. So yes, and you can watch it happen.

4:03Fifth powers though. No square to split. But the same construction works: x minus y, times five terms stepping down one at a time. I ran every power up to eight and it comes out exactly right each time. And there is a reason that costs one line. Put x equal to y, and x to the n minus y to the n is nothing. Something that vanishes when x equals y has x minus y as a factor.

4:36Try it with a plus and it fails, which is the point. Two equal numbers make that sum twice something, not nothing. Now widen it by a letter and watch the same machine run. x plus y plus z, times x squared plus y squared plus z squared, minus x y, minus x z, minus y z. Three terms against six. Eighteen products. That sounds like a mess. It is not, because the second bracket was built the same way.

5:12Twelve of the eighteen die, in six pairs. Three survive untouched: x cubed, y cubed, z cubed. And one more thing is standing, which is the interesting one. What is left over is minus three x y z. Not minus one. Three. And nobody chose that three. Among the products the x makes is x times minus y z. One minus x y z. Among the products the y makes is y times minus x z. Another.

5:47And z times minus x y. A third. Three groups, one each, and none can cancel, because nothing else in the eighteen has that shape. So x cubed plus y cubed plus z cubed minus three x y z equals x plus y plus z, times that second bracket. The three counts groups. That is all it has ever been. Now what that identity is for, because it buys something the others do not.

6:19Three numbers. Their sum is ten. Their product is twenty five. Their squares add to thirty eight. Find the sum of their cubes. The instinct is to find the three numbers first. Do not. Every fact you were given is symmetric. Shuffle the numbers and the sum, the product and the total of squares are unchanged. So is the answer. And this identity converts exactly that kind of information into a sum of cubes.

6:51Substitute. Ten, times thirty eight minus the pair sums, equals the sum of cubes minus three twenty fives. One quantity missing: x y plus y z plus z x. Which the other identity hands over. The square of a sum of three is the sum of the squares, plus twice the pair sums. So a hundred is thirty eight plus twice the pair sums. They come to thirty one. Back into the first. Ten times seven is seventy, plus seventy five.

7:27A hundred and forty five. I checked it a second way, building the power sums up without touching the identity, and it lands on the same number. Now something honest, because I looked. Those three facts pin down a cubic whose discriminant is negative. Only one of its roots is real. So no three real numbers have that sum, that product and that total of squares. The answer is still a hundred and forty five, because every step was algebra and never needed them to exist.

8:03This was never a shortcut to finding the numbers. It does not need numbers. Backwards now. p cubed, plus twenty seven q cubed, plus r cubed, minus nine p q r. Three cubes and a product term, so try the identity. Twenty seven q cubed is three q, cubed. The three slots are p, three q and r. Check the last term. Three times p times three q times r is nine p q r. It matches.

8:38So it comes apart as p plus three q plus r, times the second bracket that matches it. Harder one. Nine x cubed, minus eight thirds y cubed, plus z cubed over three, plus six x y z. Nothing there is a cube. But take a third out of everything and you get twenty seven x cubed, minus eight y cubed, plus z cubed, plus eighteen x y z. Now they are cubes: three x, minus two y, z. And three times those three is minus eighteen, so the plus eighteen is right.

9:17The common factor first. That is the whole difficulty of it. Two last things this buys. First. n cubed minus n is divisible by six, for every whole number. No rule to learn. Factorise it: n minus one, times n, times n plus one. Three consecutive numbers. Exactly one of any three is a multiple of three. At least one is even. So the product carries a two and a three, and six divides it.

9:52I ran the first five hundred. It holds every time, and so does the reason. Second, the one that looks like magic. Suppose three numbers add to nothing. Then the first bracket is nothing, so the sum of their cubes is exactly three times their product. Not a general fact. Of three hundred and forty three triples, only forty three satisfy it — exactly the ones adding to nothing, or all equal.

10:23But spot it, and an expression that looked like work is nought in one line. The bracket was built to cancel. It is still doing it.