Browse · MATH Print → jmc algebra intermediate Problem Let x,y,z be real numbers so that y+zz+xx+y=13,=14,=15. Find xyz(x+y+z). Solution — click to reveal Add the three equations together and divide by 2 to get x+y+z=21. Hence, x=8,y=7,z=6, and xyz(x+y+z)=21(8)(7)(6)=24⋅32⋅72=84. Final answer 84 ← Previous problem Next problem →