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Question

Question: If x + y + z = 12 & x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 96 &\(\frac{1}{x} + \frac{1}{y} ...

If x + y + z = 12 & x2 + y2 + z2 = 96 &1x+1y+1z\frac{1}{x} + \frac{1}{y} + \frac{1}{z}= 36

then value of x3 + y3 + z3 is

A

862

B

863

C

865

D

866

Answer

866

Explanation

Solution

(x + y + z)2 = 144

\ åx2 + 2åxy = 144

̃ åxy = 24

Now

xyxyz\frac{\sum xy}{xyz}= 36

̃ xyz = 23\frac{2}{3}

\ x3 + y3 + z3 – 3xyz = (x + y + z) (åx2 – åxy)

\ åx3 – 2 = 12 (96 – 24)

̃ åx3 = 866