Question

verify that x³+y³+z3_3xyz=1/2(x+y+z)[( x_y)²+(y_z)²+(z_x)²]​

Answer 1

Solution :

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

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

[(a-b)²=a²+b²-2ab]

⇒1/2x(x+y+z) (2x² + 2y² +22²-2xy -2yz-2xz)

1/2x(x+y+z) 2(x² + y² + 2² - xy-yz -xz)

= (x + y + 2) (x² + y² + z² - xy - yz -xz)

= x+y+z²-3xyz = L.H.S

We know that,

[x² + y² + 2²-3xyz = (x+y+z)(x² + y² + z² - xy - yz -xz)]

L.H.S = R.H.S

[x²³+ y² + 2²-3xyz

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

Its easy...