a: Đặt \(a^2+b^2=x\)
Ta có: \(M=\left(a^2+b^2+2\right)^3-\left(a^2+b^2-2\right)^3-12\left(a^2+b^2\right)^2\)
\(=\left(x+2\right)^3-\left(x-2\right)^3-12x^2\)
\(=x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)-12x^2\)
\(=x^3-6x^2+12x+8-x^3+6x^2-12x+8\)
\(=8\)
b: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1^3-3ab+3ab=1\)