<Question>
Find the quotient when $x^9 + 1$ is divided by $x - 1.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the quotient when x^9 + 1 is divided by x - 1. Hmm, let's see. I remember that when you divide a polynomial by a linear divisor like x - a, you can use polynomial long division or synthetic division. Alternatively, maybe there's a theorem or something that can help me here.

Wait, the Remainder Theorem states that if a polynomial f(x) is divided by x - a, the remainder is f(a). So if I plug in x = 1 into x^9 + 1, that should give me the remainder.
Let me check that first.

<prompt_1.1> Use the Remainder Theorem to find the remainder when \(x^9 + 1\) is divided by \(x - 1\). </prompt_1.1>

But the question asks for the quotient, not just the remainder. So how do I find the quotient?

If I recall, polynomial long division is the method where you divide one polynomial by another. So maybe I should set up the division of x^9 + 1 by x - 1.

Alternatively, maybe there's a shortcut here. Since x^9 + 1 is a sum of two terms, perhaps I can factor it. The sum of cubes formula is a^3 + b^3 = (a + b)(a^2 - ab + b^2). But here, the exponent is 9, which is 3*3, so maybe it's a sum of cubes twice?

<prompt_1.2> Consider factoring \(x^9 + 1\) using the sum of cubes formula. </prompt_1.2>
-----

Alternatively, if I perform polynomial long division, let's see how that works. Let's divide x^9 + 1 by x - 1.

<prompt_1.3> Perform polynomial long division to divide \(x^9 + 1\) by \(x - 1\). </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's compute f(1): 1^9 + 1 = 1 + 1 = 2. So the remainder is 2. That means when x^9 + 1 is divided by x - 1, the remainder is 2. But the quotient would be the polynomial resulting from the division, right? So the division can be expressed as (x^9 + 1) = (x - 1)*Q(x) + 2, where Q(x) is the quotient. Therefore, to find Q(x), I need to subtract 2 from x^9 + 1 and then divide by x - 1. So Q(x) = (x^9 + 1 - 2)/(x - 1) = (x^9 - 1)/(x - 1).
</execution_1.1>

<execution_1.2>
Let me try that.

Wait, x^9 is (x^3)^3, so x^9 + 1 = (x^3)^3 + 1^3. Applying the sum of cubes formula, that would be (x^3 + 1)(x^6 - x^3 + 1). Then, x^3 + 1 can be factored further as (x + 1)(x^2 - x + 1). So putting it all together, x^9 + 1 = (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1).
</execution_1.2>

<execution_1.3>
The dividend is x^9 + 0x^8 + 0x^7 + ... + 0x + 1, and the divisor is x - 1. 

First term: x^9 divided by x gives x^8. Multiply the divisor by x^8: x^8*(x - 1) = x^9 - x^8. Subtract this from the dividend:

(x^9 + 0x^8 + ... ) - (x^9 - x^8) = x^8. Bring down the next term, which is 0x^7. 

Next term: x^8 divided by x is x^7. Multiply divisor by x^7: x^7*(x - 1) = x^8 - x^7. Subtract:

(x^8 + 0x^7) - (x^8 - x^7) = x^7. Bring down the next term, which is 0x^6. 

This pattern will continue, right? Each time, the remainder after subtraction is x^n, and then we multiply by x^{n-1} and subtract, leading to the next term. So after each step, the quotient will be x^8 + x^7 + x^6 + ... + x + 1. 

When we get to the constant term, the last step: we have 1 (from the dividend) minus (-1)*1 (from the divisor multiplied by the last term of the quotient, which would be 1). So 1 - (-1) = 2, which is the remainder. So again, the quotient is x^8 + x^7 + ... + x + 1, and the remainder is 2. 

Therefore, using both polynomial long division and the factorization method, we arrive at the same quotient. So I think that's the answer.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
But how does that help me find the quotient when divided by x - 1?

Alternatively, maybe I can use the fact that x^9 - 1 factors as (x - 1)(x^8 + x^7 + ... + x + 1). Wait, yes, that's a standard formula.

<prompt_2.1> Use the standard formula for factoring \(x^9 - 1\) to find the quotient. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
For any n, x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + x + 1). So if n is 9, then x^9 - 1 = (x - 1)(x^8 + x^7 + ... + x + 1). Therefore, (x^9 - 1)/(x - 1) = x^8 + x^7 + ... + x + 1. 

But in our original problem, we have (x^9 + 1)/(x - 1). Wait, x^9 + 1 is not x^9 - 1. But earlier, using the Remainder Theorem, we saw that the remainder is 2. So (x^9 + 1)/(x - 1) = Q(x) + 2/(x - 1), where Q(x) is the quotient. But according to the factoring approach, x^9 + 1 can be written as (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1). Hmm, but how does that relate to division by x - 1?
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Maybe if I factor x^9 + 1 as (x + 1)(x^8 - x^7 + x^6 - ... + x^2 - x + 1). Wait, is that right?

<prompt_3.1> Factor \(x^9 + 1\) as a sum of cubes and find the quotient. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me check. The sum of odd powers: x^9 + 1 can be written as x^9 + 1^9. The formula for sum of odd powers: a^9 + b^9 = (a + b)(a^8 - a^7b + a^6b^2 - ... + b^8). So yes, x^9 + 1 = (x + 1)(x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1). So that's another factorization. 

But again, how does this help me divide by x - 1? If I have x^9 + 1 factored as (x + 1)(that alternating polynomial), and I want to divide by x - 1, which is a different factor. So unless x - 1 is a factor of x^9 + 1, which I don't think it is. Let's check by plugging in x = 1 into x^9 + 1: 1 + 1 = 2 ≠ 0, so x - 1 is not a factor. Therefore, when we divide x^9 + 1 by x - 1, the quotient will be a polynomial of degree 8, and the remainder is 2, as we found earlier.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_1.3:
Let me just verify once more with an example.

<prompt_4.1> Verify the solution with a smaller exponent, such as \(x^3 + 1\). </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Suppose we take x^3 + 1 divided by x - 1. Using the same method, the remainder is 1^3 + 1 = 2. Then the quotient should be x^2 + x + 1. Let's check: (x - 1)(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1. Wait, but x^3 + 1 divided by x - 1 would be (x^3 + 1)/(x - 1) = x^2 + x + 1 + 2/(x - 1). But actually, x^3 + 1 = (x + 1)(x^2 - x + 1). So if we divide by x - 1, the quotient is still x^2 + x + 1 with a remainder of 2. So in that case, yes, the quotient is x^2 + x + 1. So similarly, for x^9 + 1, the quotient should be x^8 + x^7 + ... + x + 1. 

Therefore, I think the answer is x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1.
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the quotient when $x^9 + 1$ is divided by $x - 1$ is $\boxed{x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1}$.
</Final_answer>
</think>