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 &x1+y1+z1= 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
xyz∑xy= 36
̃ xyz = 32
\ x3 + y3 + z3 – 3xyz = (x + y + z) (åx2 – åxy)
\ åx3 – 2 = 12 (96 – 24)
̃ åx3 = 866