= Solution
Eliminate the first entries of rows two, three and four using multipliers $a^3,a^2,a$. All remaining subdiagonal entries are then already zero. The unit-lower-triangular <LU decomposition> is
$$
\boxed{L=\begin{pmatrix}1&0&0&0\\a^3&1&0&0\\a^2&0&1&0\\a&0&0&1\end{pmatrix},\quad
U=\begin{pmatrix}1&a&a^2&a^3\\0&\gamma&a\gamma&a^2\gamma\\0&0&\gamma&a\gamma\\0&0&0&\gamma\end{pmatrix}.}
$$
Multiplication verifies $LU=A$. Since $\gamma\ne0$, all pivots are nonzero. Forward substitution in $Ly=b$ gives
$$
y=(\gamma,-a^3\gamma,-a^2\gamma,0)^T.
$$
Backward substitution in $Ux=y$ gives $x_4=0$, $x_3=-a^2$, $x_2=0$, and $x_1=\gamma+a^4=1$. Thus
$$
\boxed{x=(1,0,-a^2,0)^T.}
$$
Direct substitution yields $Ax=(\gamma,0,0,a\gamma)^T$ as a check.
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