If the equations  x3 + 10x2 + ax + b  =  0  and  x3 + 4x2 + ax + c  =  0  have two roots in common, then the product of uncommon roots of two equations is:

The given equations are:

x³ + 10x² + ax + b = 0 ----(1)

x³ + 4x² + ax + c = 0 ----(2)

Given that these two equations have two roots in common, let's denote these common roots as p and q. The third root of equation (1) is r and the third root of equation (2) is s.

According to Vieta's formulas, the sum of the roots of a cubic equation ax³ + bx² + cx + d = 0 is -b/a, and the product of the roots taken three at a time is -d/a.

Applying this to equation (1), we have:

p + q + r = -10 ----(3)

pqr = -b ----(4)

Applying this to equation (2), we have:

p + q + s = -4 ----(5)

pqs = -c ----(6)

Subtracting equation (5) from equation (3), we get:

r - s = -6 ----(7)

Dividing equation (6) by equation (4), we get:

s/r = c/b ----(8)

From equations (7) and (8), we can solve for r and s:

r = -6s/(1 - c/b)

s = -6r/(1 - b/c)

The product of the uncommon roots r and s is rs = 36/(1 - c/b)(1 - b/c) = 36/(bc - 1).