Question

Verify that : X to the power 3 + y to the power 3 + z to the power 3-3xyz=1\2(x+y+z)[(x-y)to the power 2+(y-z) to the power 2+(z-x) to the power 2

Answer 1

Dear student,

RHS=1/2[(x+y+z){(x-y)² + (y-z)² +(z-x)² }]


Taking [(x-y)² +(y-z)² +(z-x)² ]

Expanding:
x²-2xy+y²+y²-2yz+z²+z²-2-z+x²
=2(x²+y²+Z²-xy-yz-xz)

Multiplying with (x+y+z)

=(x+y+z)[(x-y)²+(y-z)²+(z-x)²}]

=2(x+y+z)(x²+y²+z²-xy-yz-xz)

=2(x³+xy²+xz²-x²y-xyz-x²z+
​​​​​​y²x+y³+z²y-xy²-y²z-xyz
+x²z+y²z+z³-xyz-yz²-z²x)

We can see that all the remaining terams other than the following will cancel out as there are both +ve and -ve values: we get;

=2(x³+y+z³-3xyz)

Dividing by 1/2
=(X³+y³+z³-3xyz)=1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]=LHS.
HENCE PROVED.
thank you.