<s>[INST] 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
 [/INST] 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.
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