JEE Novice - (22)

Algebra Level 3

{x+y1+z=1z+z2x2xy+y2xy3z=z2+3z+9x2+xy+y2\large \begin{cases} \dfrac{x+y}{1+z} & = & \dfrac{1-z+z^2}{x^2-xy+y^2} \\ \dfrac{x-y}{3-z} & = & \dfrac{z^2+3z+9}{x^2+xy+y^2} \end{cases}

If x,y,zx,y,z satisfy the system of equations above, find the value of z3y3z^3-y^3.


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