Browse · MATH Print → jmc algebra senior Problem Let a and b be nonzero complex numbers such that a2+ab+b2=0. Evaluate (a+b)9a9+b9. Solution — click to reveal Since a2+ab+b2=0, (a−b)(a2+ab+b2)=0. This simplifies to a3−b3=0, so a3=b3.Then b9=a9. Also, (a+b)2=a2+2ab+b2=(a2+ab+b2)+ab=ab,so (a+b)3=ab(a+b)=a(ab+b2)=a(−a2)=−a3.Then (a+b)9=(−a3)3=−a9, so (a+b)9a9+b9=−a92a9=−2. Final answer -2 ← Previous problem Next problem →