a: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=54\)
hay x=1
b: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
hay x=-1
a. (x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) - 3x2 = 54
<=> x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 - 4x2 + 2x + 4x2 - 2x + 1 - 3x2 = 54
<=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 - 4x2 + 2x + 4x2 - 2x + 1 - 3x2 = 54
<=> x3 - 9x3 + 8x3 + 9x2 - 6x2 - 4x2 + 4x2 - 3x2 + 27x - x + 2x - 2x = 26
<=> 26x = 26
<=> x = 1