The product of real roots of the equation $|x{|}^{6/5}-26|x{|}^{3/5}-27=0$ is

Put $|x{|}^{3/5}=y$ [See Theory], so that (1) becomes

$\begin{array}{l}{y}^2-26y-27=0\\ \Rightarrow (y+1)(y-27)=0\Rightarrow y=-1\end{array}$

or 27

As $y=|x{|}^{3/5}\ge 0\mathrm{\forall }x$ we get y=27

$\Rightarrow |x{|}^{3/5}=3^3\Rightarrow |x|=3^5\Rightarrow x=\pm 3^5$

$\therefore$ product of roots of (1) is $\left(3^5\right)\left(-3^5\right)=-3^{10}$