Question

X-xy+z-zy/1-3y+3y2-y3

Answer 1

= х(1-у)+z(1-у)/1-у*3-3у(1-у)=(х+z)(1-у)/(1-у)(1+у+у*2)-3у(1-у)=(х+z)(1-у)/(1-у)(1+у+у*2-3у)=х+z/1-2у+у*2