Question

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

Answer 1

Answer:

Step-by-step explanation:

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.

I hope it helps you

Read more on Brainly.in - https://brainly.in/question/3983276#readmore