Official solution
\[{e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0\]
\[ \Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}\]
\[{\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12\]
\[ \Rightarrow \underbrace {{e^x} - {e^{ - x}} - {e^{ - 2x}}}_{increa{\mathop{\rm sing}\nolimits} \,(let\,f(x))} = {{12} \over {\underbrace {{e^{3x}} - 1}_{decrea{\mathop{\rm sing}\nolimits} \,(let\,g(x))}}}\]
\[\Rightarrow\] No. of real roots = 2