The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of  $(x+3)^4 + (x+5)^4 = 16$ is
                            (a) 0            (b) 2
                            (c) 4            (d) none of these

Solution:     put $y = x + (3+5)/2 =  x+4$
    the equation becomes

=>  $(y-1)^4  + (y+1)^4  =  16$ ---- (i)
  => $2{y^4 + 6(y)^2 + 1 } =  16$ --------(ii)      
My question is how was (i) converted to (ii)? I just couldn't get it. Please help?

Multiply: $(y-1)^4-(y+1)^4=(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)$. Add.
