if A is singular the A^3+A^2+A is singular
You should have learned that det(AB)= det(A)det(B) (as TheEmptySet mentioned). If AB= I then det(A)det(B)= 1 so neither det(A) nor det(B) can be 0.
However, Ackbeet's suggestion gives a more fundamental proof than that. If $\displaystyle A^3+ A^2+ A$ had an inverse, then there would exist some matrix, B, such that $\displaystyle (A^3+ A^2+ A)B= I$ . But that is the same as $\displaystyle A[(A^2+ A+ I)B]= I$ which says that $\displaystyle (A^2+ A+ I)B$ is inverse to A contradicting the fact that A does not have an inverse.