Question

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

Answer 1

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.

.

I have send a picture solution also.

.

Hope the answer helped you