（一）因a³-3a²+5a-3=(a³-a²)-(2a²-2a)+(3a-3)=a²(a-1)-2a(a-1)+3(a-1)=(a-1)(a²-2a+3)=(a-1)[(a-1)²+2].故a³-3a²+5a=3.===>(a-1)[(a-1)²+2]=0.===>a=1.(二）LZ,若是初中題,可能是b³-3b²+5b=6.===>b³-2b²-b²+2b+3b-6=0.===>b²(b-2)-b(b-2)+3(b-2)=0.===>(b-2)(b²-b+3)=0.===>(b-2)[(b-0.5)²+(11/4)]=0.===>b=2.故a+b=3.【注：若b³-3b²+5b=5.構(gòu)造函數(shù)f(x)=x³-3x²+5x-5.求導(dǎo)f′(x)=3x²-6x+5=3(x-1)²+2＞0.故在R上,函數(shù)f(x)遞增,又f(2)=1>0,f(3/2)=-7/8