27x^3+27x^2+9x+1-(1+3x)^2=0 tìm x mong các anh chị giúp giùm e ạ

`⇒27x^3+27x^2+9x+1-(1+3x)^2=0`

`⇒27x^3+27x^2+9x+1-1+6x-9x^2=0`

`⇒27x^3+18x^2+3x=0`

`⇒3x(9x^2+6x+1)=0`

$⇒\left[ \begin{array}{l}3x=0\\9x^2+6x+5=0\end{array} \right.$

$⇒\left[ \begin{array}{l}x=0\\(3x+1)^2=0\end{array} \right.$

$⇒\left[ \begin{array}{l}x=0\\x=-\frac{1}{3}\end{array} \right.$

Vậy `x=0; x=-1/3`