Okay, so I need to find the ordered triple (p, q, r) such that the polynomial x³ - 3x² + 4x - 1 is a factor of x⁹ + p x⁶ + q x³ + r. Hmm, let's start by recalling that if a polynomial f(x) is a factor of another polynomial g(x), then all the roots of f(x) must also be roots of g(x). So, if α is a root of x³ - 3x² + 4x - 1, then it must satisfy x⁹ + p x⁶ + q x³ + r = 0. 

Let me denote the given polynomial as f(x) = x³ - 3x² + 4x - 1. Since it's a cubic, it has three roots, say α, β, γ. Then, for each of these roots, we have f(α) = 0, f(β) = 0, f(γ) = 0. Therefore, substituting each root into the polynomial x⁹ + p x⁶ + q x³ + r must yield zero. So, we have equations like α⁹ + p α⁶ + q α³ + r = 0, and similarly for β and γ. 

Since we need to find p, q, r such that these equations hold for all roots α, β, γ, maybe we can use the fact that the given polynomial divides the other polynomial, and so we can perform polynomial division. Alternatively, we can express x⁹ in terms of f(x) to reduce the degree. Let me try that approach.

Given that f(x) = x³ - 3x² + 4x - 1, we can express x³ as 3x² - 4x + 1. Therefore, any higher power of x can be reduced modulo f(x). For example, x⁴ = x * x³ = x*(3x² - 4x + 1) = 3x³ - 4x² + x. But since x³ can be replaced by 3x² - 4x + 1, substitute that into the expression: 3*(3x² - 4x + 1) - 4x² + x = 9x² - 12x + 3 - 4x² + x = 5x² - 11x + 3. So, x⁴ = 5x² - 11x + 3.

Similarly, we can compute higher powers step by step. Let's try to compute x⁶ and x⁹ in terms of lower-degree polynomials to find a pattern or a recurrence relation.

Let me start by computing x³ = 3x² - 4x + 1.

Then x⁴ = x * x³ = x*(3x² -4x +1) = 3x³ -4x² +x. Substitute x³:

3*(3x² -4x +1) -4x² +x = 9x² -12x +3 -4x² +x = 5x² -11x +3.

Similarly, x⁵ = x * x⁴ = x*(5x² -11x +3) = 5x³ -11x² +3x. Substitute x³:

5*(3x² -4x +1) -11x² +3x = 15x² -20x +5 -11x² +3x = 4x² -17x +5.

x⁶ = x * x⁵ = x*(4x² -17x +5) = 4x³ -17x² +5x. Substitute x³:

4*(3x² -4x +1) -17x² +5x = 12x² -16x +4 -17x² +5x = -5x² -11x +4.

x⁷ = x * x⁶ = x*(-5x² -11x +4) = -5x³ -11x² +4x. Substitute x³:

-5*(3x² -4x +1) -11x² +4x = -15x² +20x -5 -11x² +4x = -26x² +24x -5.

x⁸ = x * x⁷ = x*(-26x² +24x -5) = -26x³ +24x² -5x. Substitute x³:

-26*(3x² -4x +1) +24x² -5x = -78x² +104x -26 +24x² -5x = (-78x² +24x²) + (104x -5x) -26 = -54x² +99x -26.

x⁹ = x * x⁸ = x*(-54x² +99x -26) = -54x³ +99x² -26x. Substitute x³:

-54*(3x² -4x +1) +99x² -26x = -162x² +216x -54 +99x² -26x = (-162x² +99x²) + (216x -26x) -54 = -63x² +190x -54.

So, x⁹ = -63x² +190x -54.

Therefore, the original polynomial x⁹ + p x⁶ + q x³ + r can be expressed in terms of lower-degree polynomials. But we need to substitute x⁹, x⁶, x³ with expressions in terms of x², x, and constants. Wait, but x⁶, x³, and x⁹ are already expressed in terms of x², x, and constants. Let me write down all the expressions:

x³ = 3x² -4x +1

x⁶ = -5x² -11x +4

x⁹ = -63x² +190x -54

So, substituting back into the original polynomial:

x⁹ + p x⁶ + q x³ + r = (-63x² +190x -54) + p*(-5x² -11x +4) + q*(3x² -4x +1) + r.

Now, let's expand this expression:

First, distribute the p and q:

= -63x² +190x -54 + (-5p x² -11p x +4p) + (3q x² -4q x +q) + r

Now, combine like terms:

x² terms: (-63 -5p +3q) x²

x terms: (190 -11p -4q) x

constant terms: (-54 +4p + q + r)

Since the original polynomial is supposed to be divisible by f(x), the remainder when divided by f(x) must be zero. Therefore, all coefficients of x², x, and the constant term must be zero. So, we can set up the following system of equations:

1. Coefficient of x²: -63 -5p +3q = 0

2. Coefficient of x: 190 -11p -4q = 0

3. Constant term: -54 +4p + q + r = 0

So, we have three equations with three unknowns p, q, r. Let's write them down:

Equation 1: -5p + 3q = 63

Equation 2: -11p -4q = -190

Equation 3: 4p + q + r = 54

Now, we need to solve for p, q, r. Let's start with Equations 1 and 2 to find p and q.

Equation 1: -5p + 3q = 63

Equation 2: -11p -4q = -190

Let's solve these two equations. Maybe use elimination. Let's multiply Equation 1 by 4 and Equation 2 by 3 to eliminate q:

Equation 1 *4: -20p +12q = 252

Equation 2 *3: -33p -12q = -570

Now, add the two equations:

(-20p -33p) + (12q -12q) = 252 -570

-53p = -318

Therefore, p = (-318)/(-53) = 318/53. Let's compute that.

Divide 318 by 53: 53*5=265, 318-265=53. So, 53*6=318. Therefore, p=6.

Wait, 53*6=318? Let's check: 50*6=300, 3*6=18, so 300+18=318. Yes, correct. So p=6.

Now, substitute p=6 into Equation 1:

-5*6 +3q =63 => -30 +3q=63 => 3q=93 => q=31.

So, q=31.

Now, substitute p=6 and q=31 into Equation 3:

4*6 +31 + r =54 =>24 +31 +r=54 =>55 +r=54 =>r=54-55= -1.

Therefore, r= -1.

So, the ordered triple is (6, 31, -1). Let me verify this solution.

First, check Equation 1: -5*6 +3*31= -30 +93=63. Correct.

Equation 2: -11*6 -4*31= -66 -124= -190. Correct.

Equation 3: 4*6 +31 + (-1)=24 +31 -1=54. Correct.

Therefore, the solution is p=6, q=31, r=-1. Therefore, the ordered triple is (6,31,-1).

Alternatively, maybe I can check by substituting x³ =3x² -4x +1 into x⁹. Let's see:

We have x⁹ = -63x² +190x -54 as computed earlier. Then, x⁶ = -5x² -11x +4, x³ =3x² -4x +1.

So, the polynomial x⁹ +6x⁶ +31x³ -1 = (-63x² +190x -54) +6*(-5x² -11x +4) +31*(3x² -4x +1) -1.

Compute each term:

-63x² +190x -54

+6*(-5x² -11x +4) = -30x² -66x +24

+31*(3x² -4x +1) =93x² -124x +31

-1

Now, combine all terms:

x² terms: (-63 -30 +93) =0

x terms:190 -66 -124 =0

constants: -54 +24 +31 -1=0

Yes, all coefficients are zero, so the polynomial x⁹ +6x⁶ +31x³ -1 is indeed divisible by x³ -3x² +4x -1. Therefore, the answer is (6,31,-1).
Therefore, the ordered triple $(p, q, r)$ is $\boxed{(6, 31, -1)}$.
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