Question

cube root of (x+1)(y+2)(2+3)=7 ; find x×y×z

Answer 1

$(x+1)(y+2)(z+3)=7\\\\xyz+3xy+2xz+yz+6x+3y+2z+6=7\\\\\\x=\frac{-yz-3y-2z+1}{yz+3y+2z+6}\\\\\\\\\y=\frac{-2xz-6x-2z+1}{xz+3x+z+3}\\\\\\\\z=\frac{-3xy-6x-3y+1}{xy+2x+y+2}\\\\Hence,\\\\\:\:\:\: xyz = (\frac{-yz-3y-2z+1}{yz+3y+2z+6}) \times(\frac{-2xz-6x-2z+1}{xz+3x+z+3}) \times (\frac{-3xy-6x-3y+1}{xy+2x+y+2})$