Question

Cho a + 1 = 1. Tính:

A = a³ + b³ + 3ab. (a² + b²) + 6a²b². (a + b)

Answer 1

\(a+1=1\Rightarrow a=0\)

Ta có:

\(A=\left(a+b\right)^3+3ab\left(a^2+b^2+2ab\right)-6a^2b^2+6a^2b^2\left(a+b\right)\)

\(A=\left(a+b\right)^3+3ab\left(a+b\right)^2+6a^2b^2\left(a+b-1\right)\)

Thay \(a=0\), ta có:

\(A=b^3\)

Answer 2

Sửa đề: \(a+b=1\)

\(A=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=1.\left[\left(a+b\right)^2-3ab\right]+3ab\left(1^2-2ab\right)+6a^2b^2.1\)

\(=-3ab-6a^2b^2+6a^2b^2\)

\(=-3ab\)