Question

x3+y3+z3+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)]

Explanation: