Question

prove that (x+y)cube + (y+z)cube +(z+x) cube -3(x+y) (y+z) (z+x)​

Answer 1

Step-by-step explanation:

LHS part (x+y)3 + (y+z)3 + (z+x)3 - 3(x+y)(y+z)(z+x) =(x+y + y+z + z+x){(x+y)2 + (y+z)2 + (z+x)2 - (x+y)(y+z) - (y+z)(z+x) - (z+x)(x+y)} =2(x+y+z){x2+y2+2xy + y2+z2+2yz + z2+x2+2xz - xy-y2-xz-yz - yz-xy-z2-xz - xz-x2-yz-xy} =2(x+y+z)(x2 + y2 + z2 - xy - yz- zx) 2(x3+ y3+ z3 - 3xyz) =RHS part Proved