Consider the trigonometric equation 
$\frac{1}{{\cot }^{6}x+2\sqrt{2}\left|{\cos }^{3}x\right|}+\frac{\left|{\cos }^{3}x\right|{\left(\cos ecx\right)}^6}{\left|{\sec }^{3}x\right|+2\sqrt{2}}+\frac{2\sqrt{2}\left|{\sin }^{3}x\right|}{\left|{\tan }^{3}x\right|+\left|{\cot }^{3}x\right|}=\frac{3}{2},$

Which of the following option(s) is/are correct? 
 

Put  $2\sqrt{2}\left|{\sin }^{3}x\right|=a,\left|{\tan }^{3}x\right|=b,\left|{\cot }^{3}x\right|=c$
We get   $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{3}{2}\Rightarrow \frac{a+b+c}{b+c}+\frac{b+c+a}{c+a}+\frac{c+a+b}{a+b}=\frac{9}{2}$
$(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})=\frac{9}{2}$
Using  $AM\ge HM,$ $\frac{(a+b)+(b+c)+(c+a)}{3}\ge \frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$
$(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\ge \frac{9}{2}$
     $a=b=c 2\sqrt{2}\left|{\sin }^{3}x\right|=\left|{\tan }^{3}x\right|=\left|{\cot }^{3}x\right|$