To find :

\( \Large  \sqrt{\frac{ \left(0.064-0.008\right) \left(0.16-0.04\right) }{ \left(0.16+0.08+0.04\right) \left(0.4+0.2\right)^{3} }} \)

Let x = 0.4 ; y = 0.2

\( \Large  \sqrt\frac{ \left(x^{3}-y^{3}\right) \left(x^{2}-y^{2}\right) }{ \left(x^{2}+xy+y^{2}\right) \left(x+y\right)^{3} } \)

since \( \Large  \left(x^{3}-y^{3}\right) = \left(x-y\right) \left(x^{2}+xy+y^{2}\right) \)

Therefore, \( \Large  \sqrt \frac{ \left(x-y\right) \left(x+y\right) \left(x-y\right) }{ \left(x+y\right)^{3} } =  \frac{ \left(x-y\right)^{2} }{ \left(x+y\right)^{2} } \)

= \( \Large  \frac{x-y}{x+y} = \frac{0.4-0.2}{0.4+0.02} = \frac{1}{3} \)