gải phương trình bằng cách đặt ẩn phụ ạ

$\displaystyle ( x-1)^{5} +( x+3)^{5} =242( x+1)$
Đặt $\displaystyle y=x+1$
$\displaystyle \Rightarrow \begin{cases}
x-1=y-2 & \\
x+3=y+2 & 
\end{cases}$
Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
( y-2)^{5} +( y+2)^{5} =242y\\
\Leftrightarrow y^{5} -5y^{4} +40y^{3} -80y^{2} +80y-32+y^{5} +5y^{4} +40y^{3} +80y^{2} +80y+32=242y\\
\Leftrightarrow 2y^{5} +80y^{3} +160y-242y=0\\
\Leftrightarrow 2y^{5} +80y^{3} -82y=0\\
\Leftrightarrow 2y\left( y^{4} +40y^{2} -41\right) =0\\
\Leftrightarrow 2y\left( y^{2} +41\right)\left( y^{2} -1\right) =0\\
\Leftrightarrow 2y( y-1)( y+1)\left( y^{2} +41\right) =0\\
\Leftrightarrow \left[ \begin{array}{l l}
y=0 & \\
y-1=0 & \\
y+1=0 & \\
y^{2} +41=0 & ( vô\ lí)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
y=0 & \\
y=1 & \\
y=-1 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x+1=0 & \\
x+1=1 & \\
x+1=-1 & 
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l l}
x=-1 & \\
x=0 & \\
x=-2 & 
\end{array} \right.
\end{array}$