1. We are given $A^2=3A+\alpha I$ for a $3\times 3$ matrix $A$, and also $A^4=21A+\beta I.$

We need to find $\alpha$ and $\beta$ from these relations.

2. Compute $A^4$ using the given relation for $A^2$.

Since $A^2=3A+\alpha I,$ we square both sides: $A^4=(A^2)^2=(3A+\alpha I)^2.$

Now expand: $A^4=9A^2+6\alpha A+\alpha^2 I.$

3. Substitute again for $A^2$.

Using $A^2=3A+\alpha I,$ we get $9A^2=9(3A+\alpha I)=27A+9\alpha I.$

So, $A^4=27A+9\alpha I+6\alpha A+\alpha^2 I.$ Group the $A$ and $I$ terms: $A^4=(27+6\alpha)A+(9\alpha+\alpha^2)I.$

4. Compare with the given expression for $A^4$.

We are also told that $A^4=21A+\beta I.$

Hence, $(27+6\alpha)A+(9\alpha+\alpha^2)I=21A+\beta I.$

Comparing coefficients of $A$ and $I$: $27+6\alpha=21$ and $\beta=9\alpha+\alpha^2.$

5. Solve for $\alpha$.

From $27+6\alpha=21,$ we get $6\alpha=-6$ $\alpha=-1.$

6. Now find $\beta$.

$\beta=9(-1)+(-1)^2=-9+1=-8.$

Thus, $\alpha=-1,\qquad \beta=-8.$

7. Check the options.

• Option A: $\alpha=1$ → false
• Option B: $\alpha=4$ → false
• Option C: $\beta=8$ → false
• Option D: $\beta=-8$ → true

Therefore, the correct option is $\boxed{\text{D}}.$