Question
If x + y + z = 4, xyz = 6 & x2 +
y2 + z2 = 8 then find the value of x3 + y3 + z3.Solution
We know the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc (4)2 = 8 + 2(xy + yz + zx) (xy + yz + zx) = 8/2 = 4 a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc So, x3 + y3 + z3 = 4(8 – (4)) + 3(6) = 16 + 18 = 34
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