Take the expression in chunks:
>>
[tex]((X+A)(X-A)M+A^2M)E=((X^2-A^2)M+A^2M)E=X^2ME[/tex]
>>
[tex]\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{\frac{R+X}X}{\frac X{R-X}}+1[/tex]
[tex]\dfrac{\frac{(R+X)(R-X)}{X(R-X)}}{\frac{X^2}{X(R-X)}}=\dfrac{(R+X)(R-X)}{X^2}=\dfrac{R^2-X^2}{X^2}=\dfrac{R^2}{X^2}-1[/tex]
[tex]\implies\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{R^2}{X^2}[/tex]
>>
Note that [tex](a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab[/tex]. So
[tex]\displaystyle\frac{4AS}3\sum_{\rm cyc}\frac1{X((E+Y)^2-(E-Y)^2)}=\frac{4AS}3\sum_{\rm cyc}\frac1{4XEY}=\frac{AS}3\sum_{\rm cyc}\frac1{XEY}[/tex]
where the cyclic sum notation means
[tex]\displaystyle\sum_{\rm cyc}f(x,y,z)=f(x,y,z)+f(z,x,y)+f (y,z,x)[/tex]
In other words, we take the sum over all possible cycles of the sequence of arguments to the summand. The second square-bracketed chunk reduces to
[tex]\displaystyle\frac{AS}3\sum_{\rm cyc}\frac1{XEY}=\frac{AS}3\left(\dfrac1{XEY}+\dfrac1{EYX}+\dfrac1{YXE}\right)=\dfrac{AS}3\dfrac3{XEY}=\dfrac{AS}{XEY}[/tex]
So to recap, we've reduced the starting expression to
[tex]X^2ME\left(\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}\right)Y[/tex]
>>
[tex]\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}=\dfrac{R^2EY-ASX}{X^2EY}[/tex]
Finally, we have
[tex]X^2ME\dfrac{R^2EY-ASX}{X^2EY}Y=M(R^2EY-ASX)[/tex]
Then distributing [tex]M[/tex] and rewriting to decode the message, we have
[tex]MERRY-XMAS[/tex]
