Constructing one-ninth fraction of the $3^{5}$ design I want to construct a $3^{5-2}$ design with $I=ABC$ and $I=CDE$ as generators. The complete defining relation for this design is: $I=ABC=CDE=ABC^{2}DE=ABD^{2}E^{2}$. This is a resolution III design with $x_4=2x_1+2x_2+x_3$ and $x_5=x_1+x_2+2x_4\:({\rm mod}\:3)$.
Question
I wonder how to get $x_4=2x_1+2x_2+x_3$ and $x5=x_1+x_2+2x_4\:({\rm mod}\:3)$ in this specific problem. 
Edited

This is a question from Design and Analysis of Experiments by Douglas C. Montgomery.
 A: Allow me to begin with the qualification that algebra is not my strong suit and in my area (industrial statistics) we use fractionated three-level factorials almost never.  Is it possible there's an error in your question?  I've generated the design given the $x_k$'s and it appears that the words you give are not actually words.  I also worked out the columns given $I=ABC=CDE$ and got a different design.

Words Given the Design Construction
I went ahead and constructed your design from the $x_k$'s you give using R:
X<-as.data.frame(expand.grid(c(0:2),c(0:2),c(0:2)))
colnames(X)<-c("A","B","C")
X[,"D"] <- (2*X[,"A"] + 2*X[,"B"] + X[,"C"]) %% 3
X[,"E"] <- (X[,"A"] + X[,"B"] + 2*X[,"D"]) %% 3

When I did both
(X[,"A"]+X[,"B"]+X[,"C"]) %% 3

and
(X[,"C"]+X[,"D"]+X[,"E"]) %% 3

I get 
[1] 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1 2 0

which suggests neither $ABC$ nor $CDE$ are words.  Let $A=x_1$,...,$E=x_5$ and assume addition on $\mathbb{F}_3$.  Then given
\begin{align}
D&=2A+2B+C,\\
E&=A+B+2D,
\end{align}
we have
\begin{align}
2D &= A + B + 2C\\
E &= 2A + 2B + 2C\\
A+B+C+E &= 0\\
\end{align}
so $I=ABCE$ should be a word.  Indeed, in R
(X[,"A"]+X[,"B"]+X[,"C"]+X[,"E"]) %% 3

yields
 [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Similarly,
\begin{align}
A+B+D&=C \\
A+B+2C+D &= 0
\end{align}
so I'd expect $I=ABC^2D$ to be a word, and in R we find
(X[,"A"]+X[,"B"]+2*X[,"C"]+X[,"D"]) %% 3

yields
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

It appears that the words are 
\begin{align}
I=ABCE=ABC^2D=A^2B^2C^2E^2=A^2B^2CD^2=A^2B^2DE=ABD^2E^2=CDE^2=C^2D^2E.
\end{align}

Design Construction Given the Words
Going the other way, let $A+B+C=0$ and $C+D+E=0$ then 
\begin{align}
C&=2A+2B\\
2C&=A+B\\
E&=2C+2D = A+B+2D.
\end{align}
It is more convenient (for me, at least) to treat $A$,$B$, and $D$ as coming from the $3^3$ and solve for $C$ and $E$.  We can express $D$ as
\begin{align}
D &= 2C + 2E \\
&= A+B+2E \\
&= 2A+ 2B+C+2E
\end{align}
but since no word contains something that's not a multiple of $DE$ (or $D+E$ in my notation) it's not possible to solve for both independently of each other.
