Browse · MATH Print → jmc algebra junior Problem Two reals a and b are such that a+b=7 and a3+b3=91. Compute ab. Solution — click to reveal We have 91=a3+b3=(a+b)(a2−ab+b2)=(a+b)((a+b)2−3ab)=7⋅(49−3ab), from which ab=12. Final answer 12 ← Previous problem Next problem →