Question

verify that :
x³+y³+z³–3xyz=1/2(x+y+z) (x–y)²+(y–z)²+(z–x)​

Answer 1

x³ + y³ + z³ – 3xyz = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]

Using R.H.S

1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]

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

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

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

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

= x³ + xy² + xz² - x²y - xyz - x²z + yx² + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - xz²

= x³ + y³ + z³ - 3xyz = L.H.S

Answer 2

Answer:

y+x+z=

$3 \times \frac{1}{2}$

Step-by-step explanation:

> x² + y³ + z³ - 3xyz = 1/2( x + y + z ) (x-y)² + (y-z)² + (z-x)

> x² + y³ + z³ - 3xyz = 1/2 + x + y + z + x ²- y² + y² - z² + z - x

1. x² + y³ + z³ - 3xyz - x - y - z - x² + y² - z + x