If a3 + b3 + c3 - 3abc = 405 and (a - b)2 + (b - c)2 + (c - a)2 = 54, find the…

Concept: For any three numbers a, b and c, the expression a3 + b3 + c3 - 3abc always factorises as (a + b + c)(a2 + b2 + c2 - ab - bc - ca). Separately, the sum of squared differences (a - b)2 + (b - c)2 + (c - a)2 is always exactly twice that same quadratic expression, that is, 2(a2 + b2 + c2 - ab - bc - ca).

Combining the two, a3 + b3 + c3 - 3abc = (a + b + c) × [(a - b)2 + (b - c)2 + (c - a)2] ÷ 2. So the cube expression, the sum of the numbers and the sum of squared differences are locked to one another: knowing any two of them fixes the third, and no individual value of a, b or c is ever needed.

Application: Both quantities given in this question feed straight into that chain.

1. The second given quantity is (a - b)2 + (b - c)2 + (c - a)2 = 54.

2. That sum equals 2(a2 + b2 + c2 - ab - bc - ca), so the quadratic factor is a2 + b2 + c2 - ab - bc - ca = 54 ÷ 2 = 27.

3. The factorisation therefore reads a3 + b3 + c3 - 3abc = (a + b + c) × 27.

4. The first given quantity is a3 + b3 + c3 - 3abc = 405, so (a + b + c) × 27 = 405.

5. Dividing both sides by 27 gives a + b + c = 405 ÷ 27 = 15.

Cross-check: A concrete triple confirms that both given quantities hold together with this total.

• Take a = 8, b = 5 and c = 2, whose total is 8 + 5 + 2 = 15.

• Then (a - b)2 + (b - c)2 + (c - a)2 = 32 + 32 + (-6)2 = 9 + 9 + 36 = 54, which matches the second given quantity.

• And a3 + b3 + c3 - 3abc = 512 + 125 + 8 - 3 × 8 × 5 × 2 = 645 - 240 = 405, which matches the first given quantity.

Hence a + b + c = 15.