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Question

If x+y+z=1, xy+yz+zx=1 and xyz=1, find the value of x3+y3+z3.

A
1
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B
4
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C
2
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D
3
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Solution

The correct option is A 1
(x+y+z)(x2+y2+z2xyyzzx)=x3+y3+z33xyz
Given x+y+z=1, xy+yz+zx=1 and xyz=1
To find x2+y2+z2=x2+y2+z2+2(xy+yz+zx)
1=x2+y2+z2+(2×1)
x2+y2+z2=1+2=3
Substituting all the values in eqn (i) we get
1×(3(1))=x3+y3+z3(3×1)
1×(3+1)=x3+y3+z3+3
4=x3+y3+z3+3
x3+y3+z3=1

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