Browse · MATH Print → jmc algebra junior Problem Two reals x and y are such that x−y=4 and x3−y3=28. Compute xy. Solution — click to reveal We have 28=x3−y3=(x−y)(x2+xy+y2)=(x−y)((x−y)2+3xy)=4⋅(16+3xy), from which xy=−3. Final answer -3 ← Previous problem Next problem →