An algebra problem by Sompong Chuisurichy

Note the following:

$\begin{array} {cl} \dfrac a{|a|} & = \begin{cases} - 1 & \text{if }a < 0 \\ +1 & \text{if }a>0 \end{cases} \\ \dfrac b{|b|} & = \begin{cases} - 1 & \text{if }b < 0 \\ +1 & \text{if }b>0 \end{cases} \\ \dfrac c{|c|} & = \begin{cases} - 1 & \text{if }c < 0 \\ +1 & \text{if }c>0 \end{cases} \\ \dfrac {a+b}{|a+b|} & = \begin{cases} - 1 & \text{if }b < 0 \\ +1 & \text{if }b>0 \end{cases} \\ \dfrac {b+c}{|b+c|} & = \begin{cases} - 1 & \text{if }c < 0 \\ +1 & \text{if }c>0 \end{cases} \\ \dfrac {c+a}{|c+a|} & = \begin{cases} - 1 & \text{if }c < 0 \\ +1 & \text{if }c>0 \end{cases} \end{array}$

Therefore,

$\begin{aligned} X & = \frac a{|a|} + \frac b{|b|} + \frac c{|c|} + \frac {a+b}{|a+b|} + \frac {b+c}{|b+c|} + \frac {c+a}{|c+a|} \\ & = \frac a{|a|} + \frac {2b}{|b|} + \frac {3c}{|c|} \\ & = \begin{cases} - 6 & \text{when }a < 0,\ b < 0, \ c < 0 \\ - 4 & \text{when }a > 0, \ b < 0, \ c < 0 \\ - 2 & \text{when }a < 0, \ b > 0, \ c < 0 \\ \ \ \ 0 & \text{when }a > 0, \ b > 0, \ c < 0 \text{ and } a < 0, \ b < 0, \ c > 0 \\ + 2 & \text{when }a > 0, \ b < 0, \ c > 0 \\ + 4 & \text{when }a < 0, \ b > 0, \ c > 0 \\ + 6 & \text{when }a > 0, \ b > 0, \ c > 0 \end{cases} \end{aligned}$

There are $\boxed 7$ possible values for $X$ .