Question

x3+y3+z3-3xyz=1/2(x+y+z)[(x-ysquare)+(y-z)square+(z-xsquare)]

Answer 1

$\textbf{ Hello Mate! }$

Gare up = We have to prove the given expression. Let's do it!

Solution =

LHS

$= {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

RHS

$= \frac{1}{2} (x + y + z)( {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} ) \\ = \frac{1}{2} (x + y + z)( {x}^{2} + {y}^{2} - 2xy + {y}^{2} + {z}^{2} - 2yz + {z}^{2} + {x}^{2} - 2xz) \\ = \frac{1}{2} (x + y + z)(2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2yz - 2zx) \\ = \frac{1}{2} (x + y + z) \times 2( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ = {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

LHS = RHS

HENCE PROVED

Have great future ahead!