The segre cubic primal $X\subset P^4$ is the GIT quotient of 6 points on $P^1$. Let $M_{0,6}$ the DM compactification of the moduli of 6pointed rational curves. The Segre primal $X$ is a cubic 3fold with ten double points and there exists a natural map $M_{0,6}\to X$ that contracts 10 boundary divisors (each iso to $P^1 \times P^1$) to the singular points. Now if I blow up the singular points of $X$, since they are ordinary double points, I get a $P^1 \times P^1$ exceptional divisor over each of them. Call $\tilde{X}$ the blown up variety. Maybe it is a silly question, but it is clear that there exists an iso $\tilde{X}\cong M_{0,6}$? Why?
From what you've written you'll get a map $M_{0,6} \to \tilde{X}$ via the universal property of blowups (see Proposition II.7.14 of Hartshorne's Algebraic Geometry). In fact this map is an isomorphism.
See the paragraph after Theorem 4.2 of HarveyLloyd Philipps, Symmetry and moduli spaces for Riemann surfaces for a short argument: http://books.google.com/books?id=pvKdD74g6UEC&pg=PA164&lpg=PA164
As Steven said, there is a morphism $\overline M_{0,6} \to \tilde{X}$ because the inverse image sheaf of the ideal of double points generates a Cartier divisor. Now $\tilde{X}$ is nonsingular, so to check the fact $\overline{M}_{0,6} \to \tilde{X}$ is an isomorphism it suffices to show injectivity of the map, or show the equality of Picard numbers.
In general, the birational morphism $\overline M_{0,n} \to (\mathbb{P}^1)^n//SL_2$ is a composition of smooth blowups and Kirwan's desingularization. Consult http://arxiv.org/abs/1002.2461.

$\begingroup$ yes I see. is the fact the inverse image sheaf of the ideal of the blow up is locally free equivalent to the schematic inverse image being smooth and locally complete intersection? $\endgroup$ – IMeasy Jan 28 '13 at 8:49

$\begingroup$ no, probably it is wrong: think of the ideal sheaf of the origin in $\mathbb{A}^2$. that's a counterexample. $\endgroup$ – IMeasy Jan 28 '13 at 8:50

$\begingroup$ Add "is of codimension 1" to your conditions and it's okay. $\endgroup$ – Steven Sam Jan 28 '13 at 15:24