1)Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
=\(x^3+1^3-\left(x^3-1^3\right)=x^3+1^3-x^3+1^3=2\)
2)
a) Ta có: \(x^4-3x^3-x+3=\left(x^4-3x^3\right)-\left(x-3\right)=x^3\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(x^3-1\right)=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)=\left(x+3\right)\left(x^2-2x\right)=\left(x+3\right)x\left(x-2\right)\)
1)(x+1)(x^2-x+1)-(x-1)(x^2+x+1)
=x^3+1-(x^3-1)
=x^3+1-x^3+1
=2
2)a)x^4-3x^3-x+3
=x^3(x-3)-(x-3)
=(x^3-1)(x-3)
=(x-1)(x^2+x+1)(x-3)
b)x^3+27+(x+3)(x-9)
=x^3+27+x^2-9x+3x-27
=x^3+x^2-6x
=x(x^2+3x-2x-6)
=x[x(x+3)-2(x+3)]
=x(x-2)(x+3)