2015/003) If $X$ and $Y$ are real root for the equations $x^3+3x^2+6x+20=0$ and $y^3+6y^2+15y-2=0$, find the sum of $X$ and $Y$

Now let
$P(x) = x^3+3x^2+6x+20$ and
$G(y) = y^3+6y^2+15y-2=0$

now let us check for p(x)

to eliminate the $x^2$ term I put x -1 to get
$P(x-1) = (x-1)^3 + 3(x-1)^2 + 6(x-1) + 20$
$= x^3-3x^2+ 3x - 1 + 3x^2 - 6x + 3 + 6x - 6 + 20$
$= x^3 + 3x + 16$

now let us check for g(y)
to eliminate the $y^2$ term I put y -2 to get
$G(y-2) =(y-2)^3+6(y-2)^2+15(y-2)-2$
$=y^3 - 6y^2 + 12y - 8 + 6y^ - 24y + 24 + 15y - 30 - 2$
$=y^3 + 3y - 16$

now if we put $R(x) = x^3 + 3x - 16$ then we get
P(x-1) = R(x) 
G(y-2) = - R(-y)

or P(x) = R(x+1) and G(y) = - R(y + 2)

so both are same and odd functions so
for the zeros of P and G the 2 values shall be -ve of each other hence

X + 1 = - (Y +2) or  X + Y  = - 3

I had solved incorrectly at http://mathhelpboards.com/challenge-questions-puzzles-28/find-x-y-13744.html#post65445.