Question

please solve questions no.11... as soon as possible..​

Answer 1

Answer:

in first term y-z-x = y-(z+x) = y - (-y) = 2y

Similarly other terms , now your fraction becomes

$( \frac{2y}{2} ) {}^{3} + ( \frac{2x}{2} ) {}^{3} + ( \frac{2z}{2} ) {}^{3} \\ \\ = {x}^{3} + {y}^{3} + {z}^{3} \\ = 3xyz$

Remember when x+y+z = 0

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz \\ = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ = 0$