prove that x³+y³+z³ -3x yz= 1/2(x+y+z)+ (x-y)²+(y-z)²+(z-x)²}
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Answered by
2
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]
Hence Prooved..........
Answered by
0
Answer:
- put the value of .x=1 y=2 z=3
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