Question

x+y+z=0 show the value of (y+z)^2/3yz + (z+x)^2/3zx + (x+y)^2/3xy = 1​

Answer 1

Answer:

x3- 2x2y + 3xy2- y3+ 2x3- 5xy2 + 3x2y- 4y3

Collecting positive and negative like terms together, we get

x3+ 2x3- 2x2y + 3x2y + 3xy2- 5xy2 - y3 - 4y3

= 3x3 + x2y - 2xy2 - 5y3