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Question

If x+y+z=9 and xy+yz+zx = 23, the value of (x3+y3+z33xyz)=?

(a) 108

(b) 207

(c) 669

(d) 729

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Solution

Givenx+y+z=9and xy+yz+zx=23

We knowx3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)

But (x+y+z)2=x2+y2+z2+2xy+2yz+2zxx2+y2+z2=(x+y+z)22xy2yz2zx

So,x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)=(x+y+z)((x+y+z)23xy3yz3zx)=(x+y+z)((x+y+z)23(xy+yz+zx)=9(923×23)=9(8169)=9×12=108

(a) 108


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