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Question

If x=21,y=13 and z=32, find x3+y3+z3xyz.

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Solution

Given x=21, y=13, z=32
here x+y+z=21+13+32=0
Hence (x+y+z)=0 ……….(1)
Now,
(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)
from equation (1) z=(x+y)
then
O=x3+y3+z3+3(x+y)(yxy)(xy+x)
O=x3+y3+z3+3(x+y)(xy)
O=x3+y3+z33xyz
x3+y3+z3xyz=2xyz
Hence x3+y3+z3xyz=2(21)(13)(32)
=2(322+1)
x3+y3+z3xyz=2342+2.

1183147_1358243_ans_cf66767018ef45d699078fbe78b177bb.JPG

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