Question

12. Verify that x³+y³+z³-3xyz= 1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]​

Answer 1

$\huge\boxed{\fcolorbox{black}{red}{ANSWER}}$

Prove that ,

x³+y³+z³-3xyz

= 1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Proof ,

To Prove this identity, we need to take help of another identity ,

We know that,

x³+y³+z³-3xyz

= ( x+y+z )( x² + y² + z² - xy - yz - zx ) .......... (i)

Now , we just need to change

( x² + y² + z² - xy - yz - zx )

as the sum of square term.

So, x² + y² + z² - xy - yz - zx

= 1/2 ( 2x² + 2y² + 2z² - 2xy - 2yz - 2zx )

= 1/2 ( x² - 2xy + y² + y² - 2yz + z² + z² - 2zx + x² )

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

From (i) , we get

x³+y³+z³-3xyz

= 1/2(x+y+z)[(x-y)²+(y-z)²+(z-x)²]

Thus , Confirmed.

Answer 2

Answer:

Answer is here mate

Hope it helps