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:

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.

I hope it helps you.

laminiaduo7 and 106 more users found this answer helpful

THANKS 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.

I hope it helps you.