Just multiply it

$\begin{aligned} \text{Quadratic equation of } \space a \text{:} \space \\ a^3 + b^3 & = (a+b)((a-b)^2) \\ 38 & = 2(a^2+b^2-ab) \\ 38 & = 2((a-b)^2 + ab)\\ 19 & = (a-b)^2 + ab \\ 19 & = (a-(2-a))^2 + a(2-a) \\ 19& = 4a^2 - 8a + 2^2 + 2a - a^2\\ 19 & = -6a + 3a^2 + 4 \\ 15 & = 3a^2 - 6a \\ 0 & = 3a^2 - 6a - 15 \\ \text{By using Vieta's formula,}\\ a_1 \times a_2 & = \frac{-15}{3} \\ &= -5\\ \text{Quadratic equation of } \space b \text{:} \space \\ 19 & = (a-b)^2 + ab \\ 19 & = ((2-b)-b)^2 + (2-b)-b \\ 19& = 4+4b^2-8b+2b-b^2\\ 19 & = -6b + 3b^2 + 4 \\ 15 & = 3b^2 - 6b \\ 0 & = 3b^2 - 6b - 15 \\ \text{By using Vieta's formula,}\\ b_1 \times b_2 & = \frac{-15}{3} \\ &= -5\\ \text{Now multiply the possibility,}\\ a_1 \times a_2 \times b_1 \times b_2 & = -5 \times -5\\ & = \color{#D61F06}{\boxed{25}}\\ \end{aligned}$

$a+b=2$

$(a+b)^3=a^3+3(ab)(a+b)+b^3=8$

Since $a+b=2$ and $a^3+b^3=38$ , $(3)(ab)(2)+38=8$ , giving us $ab=-5$ .

Now, since we are given the sum and the product of two numbers, we can deduce that solving for $a$ and $b$ will give us a quadratic equation for both $a$ and $b$ . Letting the two values of $a$ and $b$ be $a_1,a_2$ and $b_1,b_2$ respectively, we get that if $a=a_1$ , $b=b_1$ and so on, so we get that $a_1b_1=a_2b_2=-5$ . So, $a_1b_1a_2b_2 = (-5)(-5) = \boxed {25}$ .

a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})

        38=2( ( a + b )^{2} - 3ab )

        19=4 - 3ab

         -5=ab

There are 2 possibilities:

    a=x , b= -5/x

    or a= -5/x , b = x

(for some x which is a solution of (x-a)(x-b)=0)

Product of all possible a's and b's =5*5=25