<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] 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.
</s>