13 If x+y +z-0, show that x3 +ys +z33xyz.

1.	13 If x+y +z-0, show that x3 +ys +z33xyz.
Answer» Given, x^3+ y^3+ z^3= 3xyz Therefore, x^3+ y^3+ z^3- 3xyz =0 This means,x^3+ y^3+ z^3- 3xyz = (x + y + z) (x^2+ y^2+ z^2- xy - yz - zx) Now if x + y + z = 0, then...... x^3+ y^3+ z^3- 3xyz = ( 0 ) (x^2+ y^2+ z^2- xy - yz - zx) . . . . . . .... ..... ... .. . . .. . ..... [ subsituting the value of x + y + z ] x^3+ y^3+ z^3- 3xyz = 0 x^3+ y^3+ z^3= 3xyz .

Answer»

Given, x^3+ y^3+ z^3= 3xyz

Therefore, x^3+ y^3+ z^3- 3xyz =0

This means,x^3+ y^3+ z^3- 3xyz = (x + y + z) (x^2+ y^2+ z^2- xy - yz - zx)

Now if x + y + z = 0, then......

x^3+ y^3+ z^3- 3xyz = ( 0 ) (x^2+ y^2+ z^2- xy - yz - zx) . . . . . . .... ..... ... .. . . .. . ..... [ subsituting the value of x + y + z ]

x^3+ y^3+ z^3- 3xyz = 0

x^3+ y^3+ z^3= 3xyz .

Discussion