Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below.

To add to the above answer, using $A^3 = 5A^2 - 6A + 5$ to simplify further $A^2(5A^2 -33A + 70) = A(5(5A^2 - 6A + 5) - 33A^2 + 70A) = A(-8A^2 + 40A + 25) = -8(5A^2 -33A + 70) + 40A^2 + 25A = 289A - 560$

I don't see a way to find the answer directly but you can certainly simplify your calculations; since $A$ commutes with itself we may factor the polynomial however we want. Write $$ A^5-27A^3+65A^2=A^2(A^3-27A+65)=A^2\left[(A^3-5A^2+6A-5)+(5A^2-33A+70)\right]. $$ Now by Cayley Hamilton you only need to compute $A^2(5A^2-33A+70)$, which cuts down on the powers of $A$ that you need to calculate directly and reduces the problem to a bunch of addition and two matrix multiplications.