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

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.
