Step 1: State the Theorem to Be Used:
The Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation.
So if we find the characteristic equation of A and replace the scalar variable by the matrix A itself, the resulting matrix equation must hold true, which will directly give the required identity.
Step 2: Form the Characteristic Equation:
The characteristic equation is $\det(A-xI)=0$, so first write out $A-xI$.
\[ A-xI=\begin{bmatrix}1-x & 2 & 3\\3 & -2-x & 1\\4 & 2 & 1-x\end{bmatrix} \]
Step 3: Expand the Determinant:
Expand along the first row and simplify term by term.
\[ (1-x)[(-2-x)(1-x)-2]-2[3(1-x)-4]+3[6-4(-2-x)] \]
\[ =(-x^3+5x-4)+(6x+2)+(12x+42)=-x^3+23x+40 \]
Setting this to 0 and multiplying by $-1$ gives the characteristic equation $x^3-23x-40=0$.
Step 4: Apply the Theorem:
By the Cayley-Hamilton theorem, matrix A must satisfy this same equation with A in place of x and $40I$ in place of the constant term $40$.
\[ A^3-23A-40I=O \]
This is exactly the identity we were asked to show, proved here without multiplying matrices three times over.
Final Answer:
The characteristic equation of A is $x^3-23x-40=0$, so by Cayley-Hamilton theorem A satisfies it.
\[ \boxed{A^3-23A-40I=O} \]