已知4a的二此方+4a+b的平方-6a+10=0,求a的三次方b-b的三次方a

4a^2+4a+b^2-6b+10=0,
→(4a^2+4a+1)+(b^2-6b+9)=0;
→(2a+1)^2+(b-3)^2=0;
于是:只能2a+1=0且b-3=0
则a=-1/2;b=3.
则a^3·b-b^3·a
=ab·(a^2-b^2)
=ab·(a+b)(a-b).
=(-3/2)·(5/2)(-7/2)
=105/8