Question

如图，一道初二数学题

Answer 1

a^3 × b - 2 × a^2 × b^2 + a × b^3
=ab × (a^2×b - 2×a×b + b^2)
=ab × (a-b)^2
=ab × ((a+b)^2-4ab)
=3 × ((根号13)^2-4×3)
=3 × 1
=3

Answer 2

原式=ab（a-b）^2=3(（a+b）^2-4ab)=3*1=3

Answer 3

=ab(a^2-2ab+b^2)
=ab[(a+b)^2-4ab]
=3[(√13)^2-4*3]
=3

Answer 4

提出一个ab，原始变为ab（a2-2ab+b2）=ab（a-b）2，（a-b）2=（a+b）2-4ab=1，所以原式就等于3x1=3
2写后面是平方....