Align two rows only using \phantom command

Question

I want to achieve the following picture:

What I have tried so far:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\begin{matrix}
\phantom{(n+{}}2^3-1^3\phantom{{}-1)^3}&=&3\times1^2&+&3\times1&+&1\\
\phantom{(n+{}}3^3-2^3\phantom{{}-1)^3}&=&3\times2^2&+&3\times2&+&1\\
\phantom{(n+{}}4^3-3^3\phantom{{}-1)^3}&=&3\times3^2&+&3\times3&+&1\\
\phantom{(n+{}}\vdots\phantom{{}-{}}\vdots\phantom{{}-1)^3}&&\vdots&&\vdots&&\vdots\\
\phantom{(n+{}}n^3-(n-1)^3&=&3(n-1)^2&+&3(n-1)&+&1\\
(n+1)^3-n^3\phantom{{{}^3}-1)^3}&=&3n^2&+&3n&+&1\\\hline
(n+1)^3-1^3\phantom{{{}^3}-1)^3}&=&3(\sum_{i=1}^ni^2)&+&3(\sum_{i=1}^ni)&+&n\\
&=&3(\sum_{i=1}^ni^2)&+&\dfrac{3n(n+1)}{2}&+&n\\
\end{matrix}
\]
\end{document}

I think it's pretty well, however if we zoom on n^3-(n-1)^3 and (n+1)^3-n^3 we can see they are not perfectly aligned:

I can't imagine another expr in \phantom{expr} than the one I put in the code:

• Before - sign: Since the longest expression is (n+1)^3 and I have n^3, the space that compensates I think it should be (n+{} so \phantom{(n+{}}.
• After - sign: Since the longest expression is (n-1)^3 and I have n^3, the space that compensates I think it should be {{}^3}-1)^3 so \phantom{{{}^3}-1)^3}.

What should be the 2 expressions that must go in \phantom?

P.S. I know I could go for adding & before and after - sign, but I think it is a better approach using \phantom. What do you think?

Answer 1

Not the best approach. Add alignment points and let TeX do the spacing.

In the second realization I compressed the spaces and made a few cosmetic changes.

\documentclass{article}
\usepackage{amsmath,array,booktabs}

\begin{document}

\[
\begin{array}{
  @{}
  r
  @{}>{{}}c<{{}}@{}
  l
  c
  c
  c
  c
  c
  c
  @{}
}
2^3     &-& 1^3     &=& 3\times1^2 &+& 3\times1 &+& 1 \\
3^3     &-& 2^3     &=& 3\times2^2 &+& 3\times2 &+& 1 \\
4^3     &-& 3^3     &=& 3\times3^2 &+& 3\times3 &+& 1 \\[-0.6ex]
\vdots  & & \vdots  & & \vdots     & & \vdots   & & \vdots \\[0.2ex]
n^3     &-& (n-1)^3 &=& 3(n-1)^2   &+& 3(n-1)   &+& 1 \\
(n+1)^3 &-& n^3     &=& 3n^2       &+& 3n       &+& 1 \\
\midrule[1pt]
(n+1)^3 &-& 1^3     &=& 3(\sum_{i=1}^ni^2) &+& 3(\sum_{i=1}^ni)   &+& n \\
\addlinespace
        & &         &=& 3(\sum_{i=1}^ni^2) &+& \dfrac{3n(n+1)}{2} &+& n
\end{array}
\]

\[
\setlength{\arraycolsep}{0pt}
\begin{array}{
  r
  >{{}}c<{{}}
  l
  >{{}}c<{{}}
  c
  >{{}}c<{{}}
  c
  >{{}}c<{{}}
  c
}
2^3     &-& 1^3     &=& 3\times1^2 &+& 3\times1 &+& 1 \\
3^3     &-& 2^3     &=& 3\times2^2 &+& 3\times2 &+& 1 \\
4^3     &-& 3^3     &=& 3\times3^2 &+& 3\times3 &+& 1 \\[-0.6ex]
\vdots  & & \vdots  & & \vdots     & & \vdots   & & \vdots \\[0.2ex]
n^3     &-& (n-1)^3 &=& 3(n-1)^2   &+& 3(n-1)   &+& 1 \\
(n+1)^3 &-& n^3     &=& 3n^2       &+& 3n       &+& 1 \\
\midrule[1pt]
(n+1)^3 &-& 1^3 &=&
  \displaystyle 3\biggl(\,\sum_{i=1}^ni^2\biggr) &+&
  \displaystyle 3\biggl(\,\sum_{i=1}^ni\biggr) &+& n \\
\addlinespace
  && &=&
  \displaystyle 3\biggl(\,\sum_{i=1}^ni^2\biggr) &+& \dfrac{3n(n+1)}{2} &+& n
\end{array}
\]

\end{document}