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Question

If x=a+b;y=b+c;z=c+a, then the value of x3+y3+z3−3xyza3+b3+c3−3abc is equal to:

A
3
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B
2
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C
1
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D
13
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Solution

The correct option is B 2
x3+y3+z33xyz

=(x+y+z)(x2+y2+z2xyyzzx)

=(x+y+z)12[(xy)2+(yz)2+(zx)2] ....(i).

Now, x+y+z=a+b+b+c+c+a=2(a+b+c) and

xy=ac;yz=ba and zx=cb.

(i) can be written as 2(a+b+c).12[(ac)2+(ba)2+(cb)2]

=2(a+b+c)(a2+b2+c2abbcca)

=2(a3+b3+c33abc).

The value of the given expression x3+y3+z33xyza3+b3+c33abc=2.

Therefore, option B is correct.

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