a) (x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) = 28
=> x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) +(2x + 1)[(2x)2 - 2.x.1 + 12 ] = 28
=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + (2x)3 + 13 = 28
=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 + 1 = 28
=> (x3 - 9x3 + 8x3) + (9x2 - 6x2) + (27x - x) + (27 + 1) = 28
=> 3x2 + 26x + 28 = 28
=> 3x2 + 26x = 0
=> 3x2 + 26x = 0
=> \(3x\left(x+\frac{26}{3}\right)=0\)
=> 3x = 0 hoặc x + 26/3 = 0
=> x = 0 hoặc x = -26/3
b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
=> \(x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)
=> \(x^6-3x^4+3x^2-1-x^6+1=0\)
=> \(\left(x^6-x^6\right)-3x^4+3x^2+\left(-1+1\right)=0\)
=> \(-3x^4+3x^2=0\)
=> \(-\left(3x^4-3x^2\right)=0\)
=> \(3x\left(x^3-x\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x\left(x^2-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
