∵(a³+b³-c³)/(a+b-c)=c²
a³+b³-c³=ac²+bc²-c³
a³+b³=c²(a+b)
(a+b)(a²-ab+b²)=c²(a+b)
∴a²-ab+b²=c²
a²+b²=c²+ab.(1)
又由余弦定理a²+b²-c²=2abcosC得
a²+b²=c²+2abcosC.(2)
联立(1)(2)知
∴cosC=1/2
∴C=π/3
∴A+B=2π/3
又sinAsinB
=sinAsin[(2π/3)-A]
=sinA[sin(2π/3)cosA-cos(2π/3)sinA]
=(√3/2)sinAcosA+(1/2)sin²A
=(√3/4)sin2A+1/4-(1/4)cos2A
=(1/2)sin(2A-π/6)+1/4
=3/4
∴sin(2A-π/6)=1.
又∵-π/6