Consider the equation x + y + z = 15 From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) So, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz) From the question, x2 + y2 + z2 = 83 and x + y + z = 15 So, 152 = 83 + 2(xy + yz + xz) => 225 – 83 = 2(xy + yz + xz) Or, xy + yz + xz = 142/2 = 71 Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca), x3 + y3 + z3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz)) Now, x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71 So, x3 + y3 + z3 – 3xyz = 15(83 – 71) => x3 + y3 + z3 – 3xyz = 15 × 12 Or, x3 + y3 + z3 – 3xyz = 180