解:a+b=1,ab=-3
a²b+ab²=ab(a+b)=1×(-3)=-3
2a³b+2ab³=2ab(a²+b²)=2ab[(a+b)²-2ab]=2×(-3)×[1²-2×(-3)]=(-6)×7=-42
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a²b+ab² =ab(a+b) =-3 * 1 = -3
2a³b+2ab³ =2ab(a^2 +b^2) = -6 [ (a+b)^2 -2ab] = -6 * 7 = -42
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a²b+ab²=ab(a+b)=-3
2a³b+2ab³=-42
a²b+ab²=ab(a+b)=1*(-3)=-3
2a³b+2ab³=2ab(a²+b²)=2ab[(a+b)²-2ab]=2*(-3)*[1²-2*(-3)]=-6*7=-42
a^2b+ab^2=ab(a+b)=-3,2a^3b+2ab^3=-42