Question
x + y = 12, xy = 35, y + z = 16, yz = 63, z + x = 14 and
xz = 45, then find the value of x³ + y³ + z³ – 3xyz.Solution
Since, x³ + y³ + z³ – 3xyz = (x + y + z){(x + y + z)² – 3(xy + yz + zx)} So, x + y + z = (12 + 16 + 14)/2 = 21 Also, xy + yz + zx = 35 + 63 + 45 = 143 x³ + y³ + z³ – 3xyz = 21 × {21² – 3 × 143} = 21 × {441 – 429} = 21 × 12 = 252
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