We are given the matrix $A$ and need to compute $A^3 - A(a^2 + b^2 + c^2)$, where:

$A = \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix}$

Step 1: Calculate $A^2$

First, compute the square of matrix $A$ (i.e., $A \times A$):

$A^2 = \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix} \times \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix}$

Carrying out the matrix multiplication:

$A^2 = \begin{pmatrix} (c^2 + b^2) & 0 & 0 \\ 0 & (a^2 + c^2) & 0 \\ 0 & 0 & (a^2 + b^2) \end{pmatrix} = (a^2 + b^2 + c^2) I$ where $I$ is the identity matrix.

Step 2: Calculate $A^3$

Now, compute $A^3 = A \times A^2$:

$A^3 = A \times (a^2 + b^2 + c^2) I = (a^2 + b^2 + c^2) A$

Thus, $A^3 = (a^2 + b^2 + c^2) A$.

Step 3: Compute $A^3 - A(a^2 + b^2 + c^2)$

Now, subtract $A(a^2 + b^2 + c^2)$ from $A^3$:

$A^3 - A(a^2 + b^2 + c^2) = (a^2 + b^2 + c^2) A - A(a^2 + b^2 + c^2)$

Since both terms are the same, we get:

$A^3 - A(a^2 + b^2 + c^2) = 0$

Final Answer:

$A^3 - A(a^2 + b^2 + c^2) = 0$

Would you like more details or have any questions?

Here are five related questions:

1. What is the significance of the matrix $A$ in vector cross products?
2. How can we generalize this result for other matrices?
3. What role does the identity matrix play in this solution?
4. Can you explain how eigenvalues would behave for matrix $A$?
5. What are some physical interpretations of matrix $A$ in physics?

Tip: When multiplying matrices, check for symmetries and patterns; it often simplifies the calculation.