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?
 A: What you need is the expansion of $(a+b)^4$. This is a special case of the very important Binomial Theorem. We have 
$$(a+b)^4=a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2+\binom{4}{3}ab^3+b^4.$$
This simplifies to $a^4+4a^3b+6a^2b^2+4ab^3+b^4$. 
Put $a=y$, $b=1$, then $a=y$, $b=-1$, and add. There is a pleasant amount of cancellation. 
Remarks: $1.$ The symmetrizing move $y=x+4$ is a  useful idea.
$2$. For reasons of familiarity, we change the name, and study $(x+1)^4+(x-1)^4$. This function is symmetric about $x=0$. Our function is not $16$ at $x=0$, and by symmetry there are just as many solutions of $(x+1)^4+(x-1)^4=16$ with $x\gt 0$ as there are with $x\lt 0$. So let's see how many there are with $x\gt 0$.
$3.$ The solution $x=1$ is obvious. It is reasonably clear that there are no solutions with $0\lt x\lt 1$. And past $x=1$, our function is increasing. So there is exactly one positive solution, and therefore altogether there are two solutions. 
A: $$(y-1)^4+(y+1)^4$$
$$=y^4-\binom 41y^3+\binom 42y^2-\binom 43y+1+(y^4+\binom 41y^3+\binom 42y^2+\binom 43y+1)$$
$$=2(y^4+6y^2+1)$$
A: $$
\begin{align}
(y−1)^4 +(y+1)^4  
&= \left(y^4 − \binom{4}{1}y^3 + \binom{4}{2}y^2 − \binom{4}{3}y + 1\right) \\
&+ \left(y^4 + \binom{4}{1}y^3 + \binom{4}{2}y^2 + \binom{4}{3}y + 1\right) \\
&=2(y^4 + 6y^2 + 1) 
\end{align}
$$
A: Multiply: $(y-1)^4-(y+1)^4=(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)$. Add.
