Question

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

Answer 1

Đề thiếu: Cho a + b = 1. Tính giá trị biểu thức

Ta có:

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

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

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

Thay a + b = 1

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

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

\(=1\)