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Question
How (x+y+z)(x2+y2+z2-xy-yz-zx) = (x+y+z)[(x+y+z)2-3(xy+yz+zx)
How (x+y+z)(x2+y2+z2-xy-yz-zx) = (x+y+z)[(x+y+z)2-3(xy+yz+zx)
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Solution
Answer :
(x + y + z)(x2 + y2 + z2 - xy - yz - zx ) = (x + y + z)[(x + y + z)2 - 3(xy + yz + zx) ]
Taking R.H.S.
(x + y + z)[(x + y + z)2 - 3(xy + yz + zx) ]
We know
(x + y + z)2 = x 2 + y​ 2 + z2 + 2xy + 2yz + 2zx
So, we get
(x + y + z)[(x 2 + y​ 2 + z2 + 2xy + 2yz + 2zx - 3xy -3 yz - 3zx) ]
(x + y + z)(x2 + y2 + z2 - xy - yz - zx )
Hence
L.H.S. = R.H.S. ( Hence proved )
(x + y + z)(x2 + y2 + z2 - xy - yz - zx ) = (x + y + z)[(x + y + z)2 - 3(xy + yz + zx) ]
Taking R.H.S.
(x + y + z)[(x + y + z)2 - 3(xy + yz + zx) ]
We know
(x + y + z)2 = x 2 + y​ 2 + z2 + 2xy + 2yz + 2zx
So, we get
(x + y + z)[(x 2 + y​ 2 + z2 + 2xy + 2yz + 2zx - 3xy -3 yz - 3zx) ]
(x + y + z)(x2 + y2 + z2 - xy - yz - zx )
Hence
L.H.S. = R.H.S. ( Hence proved )
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