Sums and differences of cubes

Multiply (x – y)(x² + xy + y²): the middle terms cancel and you are left with x³ – y³. Put –y in place of y: (x + y)(x² – xy + y²) = x³ + y³.

A sign hook for these two: the first bracket has the same sign as the cubes; the second bracket's middle term has the opposite sign; the last term is always plus.

x – y divides every xⁿ – yⁿ:

• x² – y² = (x – y)(x + y);
• x³ – y³ = (x – y)(x² + xy + y²);
• x⁴ – y⁴ = (x² – y²)(x² + y²);
• x⁵ – y⁵ = (x – y)(x⁴ + x³y + x²y² + xy³ + y⁴).

A three-letter identity: (x + y + z)(x² + y² + z² – xy – yz – zx) = x³ + y³ + z³ – 3xyz.

Example 15. x + y + z = 10, xyz = 25, x² + y² + z² = 38. From (x + y + z)² = 100 = 38 + 2(xy + yz + zx), we get xy + yz + zx = 31. Then x³ + y³ + z³ = 3 × 25 + 10 × (38 – 31) = 145.

The special case worth remembering: if x + y + z = 0, then x³ + y³ + z³ = 3xyz. That is why x³ + y³ – 12xy + 64 is 0 when x + y = –4: it is x³ + y³ + 4³ – 3(x)(y)(4), and x + y + 4 = 0.