To solve for the remainder when \( f(x) = x^6 + x - 1 \) is evaluated at \( f(x) \) and divided by \( x + 3 \), we will use the Remainder Theorem.
Step 1: Identify the polynomial and the divisor.
Here, we want to find the remainder when \( f(f(x)) \) is divided by \( x + 3 \).
Step 2: Use the Remainder Theorem.
According to the Remainder Theorem, the remainder of \( f(x) \) divided by \( x - c \) is \( f(c) \). Thus, we need to evaluate \( f(f(-3)) \).
Step 3: Calculate \( f(-3) \).
First, compute \( f(-3) \):
\[
f(-3) = (-3)^6 + (-3) - 1
\]
Calculating step-by-step:
- \( (-3)^6 = 729 \)
- \( -3 - 1 = -4 \)
So, \[ f(-3) = 729 - 4 = 725 \]
Step 4: Evaluate \( f(f(-3)) = f(725) \).
Next, we find \( f(725) \):
\[
f(725) = 725^6 + 725 - 1
\]
Step 5: Now we find the remainder of \( f(725) \) when divided by \( x + 3 \).
Instead of calculating \( f(725) \) entirely, we must find the form it takes when divided by \( x + 3 \). By the Remainder Theorem, calculate \( f(-3) \) again but with the new value.
\[
f(x) \mod (x + 3) \ (c = -3)
\]
Using the general form of \( f(x) \):
\[
= x^6 + x - 1
\]
When evaluated at \( x = -3 \):
\[
f(-3) = 725
\]
Since \( f(x) \) is not a simple one to compute directly, instead we could find \( f(725) \mod (x + 3) \) directly by recognizing the quartic powers will cycle through mod: Let us compute directly:
- Given \( x + 3 = 0 \implies x = -3 \), we substitute this in: Thus find \( f(-3) = 725 \) again gives 725.
But as stated follow the polynomial division rules we can only find final terms when cycling:
The final answer is instead tracking it again:
Final Result:
The remainder when \( f(f(x)) \) is divided by \( x + 3 \) is \(\equiv 725 \).