ATOT Maths Challenge

[BlackTigers]

I hate factoring, lol.

 

[chuckywang]

{[x^2 + 6xy + 9y^2]/[x^2 - 9y^2]}/{[x^2 + 3xy - 2x - 6y]/[x^3 - 3yx^2 - 4x + 12y]}

[x^2 + 6xy + 9y^2]/[x^2 - 9y^2] = [(x+3y)(x+3y)]/[(x+3y)(x-3y)]

[x^2 + 3xy - 2x - 6y]/[x^3 - 3yx^2 - 4x + 12y] = [(x-2)(x+3y)]/[(x^2-4)(x-3y)

So therefore, we have (x+3y)*(x^2-4)/[(x+3y)(x-2)] = x+2 since x^2-4 = (x-2)(x+2)