Question

Verify that x³ + y³ + z³ – 3xyz 

= \(\frac{1}{2}\)(x + y + z) ](x – y)² + (y – z)² + (z – x)²]

Answer 1

L.H.S.= x³ + y³+ z³ – 3xyz 

R.H.S. = (x+y+z) [(x – y)² + (y – z)² + (z – x)²] 

= (x+y+z) [x² + 2xy + y²+ y² – 2yz + z² + z² -2zx + x² ] 

= (x+y+z) [2x²+ 2y²+ 2z²– 2xy – 2yz – 2zx] 

= (x+y+z) × (x² + y² + z² – xy – yz – zx) 

= (x+y+z)(x² + y² + z² – xy – yz – zx) 

R.H.S.= x³ + y³ + z³ – 3xyz 

∴ L.H.S. = R.H.S. 

∴ x³ + y³ + z³ – 3xyz 

= \(\frac{1}{2}\)(x + y + z)[(x – y)² + (y – z)² + (z – x)³ ]