$\int^2_0 (~~~) \times \int^2_0 (~~~)$

Note first that $\frac{\sqrt[3]a}{\sqrt a} = a^{1/3}/a^{1/2} = a^{\tfrac13 - \tfrac12} = a^{-1/6}.$ Next, $\int_0^2 \frac{\sqrt[3]{x+b}}{\sqrt{x+b}}\:dx = \int_0^2 (x+b)^{-1/6}\:dx = \left.\frac 1{5/6}(x+b)^{5/6}\right|_0^2 = \frac 6 5\left((b+2)^{5/6} - b^{5/6}\right).$ Finally, the given product of integrals evaluates to $\left(\frac 65\left(8^{5/6}-6^{5/6}\right)\right)\left(\frac 65\left(10^{5/6}-8^{5/6}\right)\right) = \frac{36}{25} \left(8^{5/6}-6^{5/6}\right)\left(10^{5/6}-8^{5/6}\right) \approx 2.0074 \approx \boxed{2.01}.$

$\begin{aligned} P & = \int_0^2 \frac {\sqrt[3]{x+6}}{\sqrt{x+6}} dx \times \int_0^2 \frac {\sqrt[3]{x+8}}{\sqrt{x+8}} dx \\ & = \int_0^2 \frac {(x+6)^\frac 13}{(x+6)^\frac 12} dx \times \int_0^2 \frac {(x+8)^\frac 13}{(x+8)^\frac 12} dx \\ & = \int_0^2 (x+6)^{-\frac 16} dx \times \int_0^2 (x+8)^{-\frac 16} dx \\ & = \frac 65 \left[(x+6)^\frac 56 \right]_0^2 \times \frac 65 \left[(x+8)^\frac 56 \right]_0^2 \\ & = \frac 65 \left[8^\frac 56 - 6^\frac 56 \right] \times \frac 65 \left[10^\frac 56 - 8^\frac 56 \right] \\ & \approx \boxed{2.01} \end{aligned}$