Taylor polynomial of $f(x_1,...,x_m)=\varphi(e^{a\sum_{i=1}^mx_i})$ 
Let $\varphi:\mathbb{R}\to\mathbb{R}$ be a $C^3(\mathbb{R})$ function, with $a\in\mathbb{R}$. Find the Taylor polynomial of degree $3$, centered in the origin $p=(0,...,0)$, of $f(x_1,...,x_m)=\varphi(e^{a\sum_{i=1}^mx_i})$.

If we define $g=e^{a\sum_{i=1}^mx_i},\ g:\mathbb{R^m}\to\mathbb{R}$, we have that $f=\varphi\circ g$, so we can calculate the Taylor polynomial of $f$ centered in $p$ (from now on $P_{3,p,f}$) by doing the compositon $P_{3,g(p),\varphi}[P_{3,p,g}]$,  where
$P_{3,p,g}=1+a\sum_{i=1}^mx_i+\frac{a^2}{2}\sum_{i,j=1}^mx_ix_j+\frac{a^3}{6}\sum_{i,j,k=1}^mx_ix_jx_k$
$P_{3,g(p),\varphi}=P_{3,1,\varphi}=\varphi(1)+\varphi'(1)(x-1)+\frac{1}{2}\varphi''(1)(x-1)^2+\frac{1}{6}\varphi'''(1)(x-1)^3$
Am I proceding in the right way? I have tried to develop last expression and replace $x$ by $a\sum_{i=1}^mx_i$, $x^2$ by $\frac{a^2}{2}\sum_{i,j=1}^mx_ix_j$ and $x^3$ by $\frac{a^3}{6}\sum_{i,j,k=1}^mx_ix_jx_k$, but it does not seem to be the right solution as it is not the same as the one in my book. What am I doing wrong? Thanks in advance!

Replacing $x$ by the polynomial $P_{3,p,g}$ (taylor series of $g(x)$) in $P_{3,g(p),\varphi}$, and ignoring the terms with degree greater than $3$, we obtain the following expression:
$\varphi(1)+\varphi'(1)(a\sum_{i=1}^mx_i+\frac{a^2}{2}\sum_{i,j=1}^mx_ix_j+\frac{a^3}{6}\sum_{i,j,k=1}^mx_ix_jx_k)+\frac{1}{2}\varphi''(1)(a^2\sum_{i,j=1}^mx_ix_j+a^3\sum_{i,j,k=1}^mx_ix_jx_k)+\frac{1}{6}\varphi'''(1)a^3\sum_{i,j,k=1}^mx_ix_jx_k=$
$\varphi(1)$+$a\varphi'(1)\sum_{i=1}^mx_i$+$\frac{a^2}{2!}(\varphi'(1)$+$\varphi''(1))\sum_{i,j=1}^mx_ix_j$+$\frac{a^3}{3!}(\varphi'(1)$+$3\varphi''(1)$+$\varphi'''(1))\sum_{i,j,k=1}^mx_ix_jx_k$
