2021/008) Given $\frac{x}{y+z} + \frac{y}{z+ x} + \frac{z}{x+y} = 1 $ Find $\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} $

Because 1st term of the LHS in the expected result has $x^2$ we need to multiply 1st term of the given expression by y , 2nd term by y and 3rd tert by z so we can multiply by x+y + z to get

$\frac{x*(x+y+z)}{y+z} + \frac{y*(x+y+z)}{z+ x} + \frac{z*(x+y+z)}{x+y} = 1 $

Or

$\frac{x^2}{y+z} + x  + \frac{y^2}{z+ x} + y + \frac{z^2}{x+y} + z = x+ y + z$

Or 

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} + x + y + z = x+ y + z$

Or

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} = 0$