已知a+b+c=0,abc=-15,求aa(b+c)+bb(a+c)+cc(a+b)

∵a+b+c=0,
∴a+b=-c b+c=-a a+c=-b
∴aa(b+c)+bb(a+c)+cc(a+b)
=-a³-b³-c³
=-a³-3a²b-3ab²-b³-c³+3a²b+3ab²
=-(a+b)³-c³+3a²b+3ab²
=-(a+b+c)(a²+2ab+b²+ac+bc+c²)+3a²b+3ab²
=-0*(a²+2ab+b²+ac+bc+c²)+3ab(a+b)
=3ab*(-c)
=-3abc
=-3*(-15)
=45