If a = 1.2, b = 2.1, c = -3.3, find the value of a3 + b3 + c3 - 3abc

Concept

There is a standard algebraic identity that factorises the sum of three cubes minus three times their product: a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca). A direct consequence is that whenever the three numbers add up to zero, i.e. a + b + c = 0, the first factor vanishes and the whole expression must equal 0 - no matter what the individual values are.

Application

1. Add the three given numbers: a + b + c = 1.2 + 2.1 + (-3.3) = 3.3 - 3.3 = 0.

2. Because the sum a + b + c equals 0, the factor (a + b + c) in the identity is 0.

3. Therefore a3 + b3 + c3 - 3abc = (0) x (a2 + b2 + c2 - ab - bc - ca) = 0.

Cross-check

Substituting directly: 1.23 = 1.728, 2.13 = 9.261, (-3.3)3 = -35.937, and 3abc = 3 x 1.2 x 2.1 x (-3.3) = -24.948. Sum of cubes = 1.728 + 9.261 - 35.937 = -24.948; subtracting 3abc gives -24.948 - (-24.948) = 0. The value is 0.