Question
x + y = 4, xy = 2, y + z = 5, yz = 3, z + x = 6 and xz
=4, then find the value of x 3 + y 3 + z 3 – 3xyz.Solution
Since, x 3 + y 3 + z 3 – 3xyz = (x + y + x){(x + y + z) 2 – 3(xy + yz + zx)}
So, x + y + z = (4 + 5 + 6)/2 = 15/2
x 3 + y 3 + z 3 – 3xyz = (15/2) × {225/4 – 3 × 9} = 15/2 × 117/4 = 1755/8
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