Find the solutions of this linear equation:

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?