Question

x^3+y^3+z^3-3xyz =1/2(x+y+z) [(x-y)^2+(y-z)^2+(z-x)^2]​

Answer 1

Answer:

First We take R.H.S & use the Formula [( a-b)²= a²+b²-2ab] & simplify it then R.H.S becomes equal to L.H.S

R.H.S

⇒ 1/2×(x + y + z) (x²+ y²-2xy +y²+ z²-2yz+x²+z²-2xz)

[( a-b)²= a²+b²-2ab]

⇒ 1/2×(x + y + z) (2x²+ 2y²+2z²-2xy -2yz-2xz)

⇒ 1/2×(x + y + z) 2(x² + y²+ z² – xy – yz – xz)

=(x + y + z) (x² + y²+ z² – xy – yz – xz)

= x³+y³+z³-3xyz

= L.H.S

We know that,

[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]

L.H.S = R.H.S

[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]

Hope this helps you