Find the square root of ( 0.064 - 0.008 ) ( 0.16 - 0.04 ) ( 0.016 + 0.08 + 0.04 ) ( 0.4 + 0.2 )^3

Answer: Option B

Square root
$$\Rightarrow \sqrt{\frac{\left(0.064-0.008\right)\left(0.16-0.04\right)}{\left(0.016+0.08+0.04\right){\left(0.4+0.2\right)}^3}}$$
$$\Rightarrow \sqrt{\frac{\left[{\left(0.4\right)}^3-{\left(0.2\right)}^3\right]\left[{\left(0.4\right)}^2-{\left(0.2\right)}^2\right]}{\left[{\left(0.4\right)}^2+{\left(0.2\right)}^2+\left(0.2\times 0.4\right)\right]{\left[\left(0.4\right)+0.2\right]}^3}}$$
$$\Rightarrow \text{Let }0.4=a,0.2=b$$
$$\Rightarrow \sqrt{\frac{\left(a^3-b^3\right)\left(a^2-b^2\right)}{\left(a^2+b^2+ab\right){\left(a+b\right)}^3}}$$
$$\Rightarrow \sqrt{\frac{\left(a-b\right)\left(a^2+b^2+ab\right)\left(a+b\right)\left(a-b\right)}{{\left(a+b\right)}^3\left(a^2+b^2+ab\right)}}$$
$$\begin{array}{rl}& \Rightarrow \sqrt{\frac{{\left(a-b\right)}^2}{{\left(a+b\right)}^2}}\\ & \Rightarrow \frac{a-b}{a+b}\\ & \Rightarrow \frac{0.4-0.2}{0.4+0.2}\\ & \Rightarrow \frac{0.2}{0.6}\\ & \Rightarrow \frac{1}{3}\end{array}$$