Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$ 
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$

$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for 
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$
$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.
$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$
So Solutions of the above equation are $x=6$ and $x=8$
Can we solve it using Derivative or Algebraic way 
Thanks
 A: I guess by Algebraic you mean this way
$$(6-x)^4+(8-x)^4 = 16$$
put $7-x=t$
(reason behind this substitution is not just because the later is easy to expand using binomial formula but the magic it does).
$$(t-1)^4+(t+1)^4 = 16$$
$$(t^4-4t^3+6t^2-4t+1)+(t^4+4t^3+6t^2+4t+1)=16$$
$$2(t^4+6t^2+1)=16$$
$$t^4+6t^2+1=8$$
Well, its quartic but it's easy to find it's roots
put $p=t^2$
$$ p^2+6p-7=0$$
$$\implies p=1,p-7$$
Now $p=-7$ will not yield any real solution so, we ignore that
$$\implies t=\pm1$$
$$x=7-t\implies x=6,8$$
Hence, only two real roots, other two being complex.
A: looking at the solution of $(6-x)^4 + (8-x)^4 = 16$ geometrically as the solution of the simultaneous equations $(6-x)^4 + (8-y)^4 = 16$ and $y = x.$ one is "squashed" circle centered at $(6,8)$ with radius $2$ and the line $y = x$ which is unit distance from the center. i expect the line to cut the circle at two points. the $x$-coordinates of those points are your two solutions. i will confirm
this when i get time to graph these.
i see from the comments that there are two solutions.
A: expanding and factorizing we obtain
$2\, \left( x-6 \right)  \left( x-8 \right)  \left( {x}^{2}-14\,x+56
 \right) 
=0$
from here you will get all solutions.
A: You can exploit the symmetry of the situation to simplify things if you set $y=7-x$ so that the equation becomes $(y-1)^4+(y+1)^4=16$ or $2y^4+12y^2+2=16$ or $y^4+6y^2-7=0$
This factorises as $(y^2-1)(y^2+7)=0$ with real solutions $y=\pm 1$ corresponding to $x=6, x=8$

Note this symmetric trick is bound to work, because it gives a symmetric function in $y$, which is inevitably a function of $y^2$ (no odd order terms) - and the order is such that it can't be worse than quadratic in $y^2$, which can therefore be solved.
A: HINT:
Write $8-x = 2 a$, $\ x-6=2b$. Then $\ a+b=1$ and $a^4+b^4=1$.  Draw some level curves. 
$\bf{added:}$

