Question

If x+y+z=0 show that x^3+y^3+z^3=3xyz

Answer 1

Hello friends!!

Here is your answer :

Given

$x + y + z = 0$


To show


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


Solution:

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$


${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (0)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$


${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0$


${x}^{3} + {y}^{3} + {z}^{3} = 0 + 3xyz$


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


Hope it helps you.. ☺️☺️☺️

Answer 2

x+y+z=0 ---------- (1) -------------------------------------------------------------squaring both sides in equation (1) (x+y+z)*3=(0)*3. x*3 + y*3 + z*3 - 3xyz =0. x*3 + y*3 + z*3 = 3xyz. LHS = RHS. ------------------------------------------------------------- verified. -------------------×--×-----×--×--------------------------