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?

$$
\begin{align}
(y−1)^4 +(y+1)^4  
&= \left(y^4 − \binom{4}{1}y^3 + \binom{4}{2}y^2 − \binom{4}{3}y + 1\right) \\
&+ \left(y^4 + \binom{4}{1}y^3 + \binom{4}{2}y^2 + \binom{4}{3}y + 1\right) \\
&=2(y^4 + 6y^2 + 1) 
\end{align}
$$
