Just use partial fractions. Write
$$
\frac{1}{(1-t)^3(1+t)(1+t+t^2)}=\frac{A_1}{1-t}+\frac{A_2}{(1-t)^2}+\frac{A_3}{(1-t)^3}+\frac{B}{1+t}+\frac{C_0+C_1t}{1+t+t^2}.
$$
Multiplying LHS by $1+t$ and letting $t=-1$ yields
$$
B=\left[\frac{1}{(1-t)^3(1+t+t^2)}\right]_{t=-1}=\frac{1}{(1+1)^3(1-1+1)}=\frac{1}{8}.
$$
Multiplying LHS by $(1-t)^3$ and letting $t=1$ yields
$$
A_3=\left[\frac{1}{(1+t)(1+t+t^2)}\right]_{t=1}=\frac{1}{(1+1)(1+1+1)}=\frac{1}{6}.
$$
You can keep going like this (using your favorite method of partial fraction expansion) to obtain the following:
\begin{multline*}
\frac{1}{(1-t)^3(1+t)(1+t+t^2)}=\\
=\frac{1}{6}\cdot\frac{1}{(1-t)^3}+\frac{1}{4}\cdot\frac{1}{(1-t)^2}+\frac{17}{72}\cdot\frac{1}{1-t}+\frac{1}{8}\cdot\frac{1}{1+t}+\frac{1}{9}\cdot\frac{2+t}{1+t+t^2}.
\end{multline*}
Everything except the last summand is expands easily into a power series, and for the last summand we can write
$$
\frac{2+t}{1+t+t^2}=\frac{(1-t)(2+t)}{(1-t)(1+t+t^2)}=\frac{2-t-t^2}{1-t^3}=\frac{2}{1-t^3}-\frac{t}{1-t^3}-\frac{t^2}{1-t^3}.
$$
Thus, noticing that $\dfrac{17}{72}=\dfrac{1}{8}+\dfrac{1}{9}$, we have
$$
\begin{split}
[t^n]\frac{1}{(1-t)^3(1+t)(1+t+t^2)}
&=\frac{1}{6}\binom{n+2}{2}+\frac{1}{4}\binom{n+1}{1}+\frac{17}{72}+\frac{(-1)^n}{8}+\frac{-1+[3 \text{ if } 3\mid n]}{9}\\
&=\frac{n^2+6n+5+3[2\mid n]+4[3\mid n]}{12},
\end{split}
$$
where "$[\cdot]$" is the Iverson bracket (i.e. $1$ if the statement in the bracket is true, and $0$ if it is false).
In this case, $2\mid 42$ and $3\mid 42$, so $[2\mid 42]=1$ and $[3\mid 42]=1$, and thus
$$
\begin{split}
[t^{42}]\frac{1}{(1-t)^3(1+t)(1+t+t^2)}&=\frac{42^2+6\cdot 42+5+3\cdot 1+4\cdot 1}{12}\\
&=\frac{42^2+6\cdot 42+12}{12}\\
&=\frac{42\cdot 48+12}{12}=42\cdot4+1=169.
\end{split}
$$
