\(A=cos^2x+\dfrac{1+cos\left(\dfrac{2\pi}{3}+2x\right)}{2}+\dfrac{1+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+\dfrac{cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+cos\left(\dfrac{2\pi}{3}\right).cos2x\\ =cos^2x+1-\dfrac{1}{2}.cos2x=\dfrac{1+cos2x}{2}+1-\dfrac{cos2x}{2}=\dfrac{3}{2}.\)

\(sin^6\left(\pi+x\right)=sin^6x,cos^6\left(x-\pi\right)=cos^6\pi\\ sin^4\left(x+2\pi\right)=sin^4x,sin^4\left(x-\dfrac{3\pi}{2}\right)=cos^4x,cos^2\left(x-\dfrac{\pi}{2}\right)=sin^2x.\)

Khi đó \(A=sin^6x+cos^6x-2sin^4x-cos^4x+sin^2x\\ =\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-\left(sin^4x+cos^4x\right)-sin^4x+sin^2x\\ =1-3sin^2x.cos^2x-\left[1-2sin^2x.cos^2x\right]-sin^2x.\left(sin^2x-1\right)\\ =1-3sin^2x.cos^2x-1+2sin^2x.cos^2x+sin^2x.cos^2x\\ =0\)

\(\left(x-1\right)^3-x^3+3x^2-3x-1\)

\(=\left(x-1\right)^3-\left(x^3-3x^2+3x-1\right)-2\)

\(=\left(x-1\right)^3-\left(x-1\right)^3-2\)

\(=-2\) (ko phụ thuộc x)

 a,2x(3x-1)-6x(x+1)-(3-8x)

=6x^2-2x-6x^2-6x-3+8x

=-3

Vậy............

Ta có 2x(3x - 1) - 6x(x  +1) - (3 - 8x) 

= 6x² - 2x - 6x² - 6x - 3 + 8x 

= (6x² - 6x²) + (8x - 2x - 6x) - 3 = - 3

=> Biểu thức trên không phụ thuộc vào biến