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

HINT:
Write $8-x = 2 a$, $\ x-6=2b$. Then $\ a+b=1$ and $a^4+b^4=1$.  Draw some level curves. 
$\bf{added:}$

