Question

Verify that x³+y³+z³-3xyz = ½(x+y+z) [(x-y)² + (y-z)²+(z-x)²]

Answer 1

HEY DEAR ... ✌️

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Here's ur answer :-

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, proved .

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HOPE , IT HELPS ... ✌️

Answer 2

The solution is provided in the attachment.