If x+y+z=0, then show that x3+y3+z3= 3xyz
ishiverma:
i hope my answer will help u cause that was easiest way of solving it.
Answers
Answered by
15
x+y+z=0
then
(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)
(0)^2 = x^2+y^2+z^2+2(xy+yz+zx)
now
x^3+y^3+z^3-3xyz= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
= 0 x 0
x^3+y^3+z^3-3xyz=0
x^3+y^3+z^3= 3xyz
hence proved
i hope it is helpful for you
then
(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)
(0)^2 = x^2+y^2+z^2+2(xy+yz+zx)
now
x^3+y^3+z^3-3xyz= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)
= 0 x 0
x^3+y^3+z^3-3xyz=0
x^3+y^3+z^3= 3xyz
hence proved
i hope it is helpful for you
Answered by
7
Hello
using 8 identity
x+y+z(x2+y2+z2_ xy_ yz_ zx)=x3 +y3 +z3 _3xyz
0=x3 + y3+ z3 _3xyz
0 = x3 + y3+z3
or. 3xyz = x3 + y3 + z3
※proved.
using 8 identity
x+y+z(x2+y2+z2_ xy_ yz_ zx)=x3 +y3 +z3 _3xyz
0=x3 + y3+ z3 _3xyz
0 = x3 + y3+z3
or. 3xyz = x3 + y3 + z3
※proved.
Similar questions