\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\Leftrightarrow6x^2+2=6x^2+6x+6\Leftrightarrow6x=-4\Leftrightarrow x=\dfrac{-2}{3}\)
\(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x+1\right)=6\left(x^2+x+1\right)\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\)
\(\Leftrightarrow-6x=4\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy tập nghiệm: \(S=\left(\dfrac{-2}{3}\right)\)
\(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(S=\left\{-\dfrac{2}{3}\right\}\)
*Cách khác:
Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]=6\left(x^2+x+1\right)\)
\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)=6\left(x^2+x+1\right)\)
\(\Leftrightarrow2\left(3x^2+1\right)=6\left(x^2+x+1\right)\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x+6-2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow6x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)
