DeepSeek vs Qwen for Math

Alright, let's work through this step-by-step.

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Step 1 – Solve $x^3 - 6x^2 + 11x - 6 = 0$ for integer solutions.

We look for integer roots using the Rational Root Theorem: possible integer roots are divisors of the constant term $-6$: $\pm 1, \pm 2, \pm 3, \pm 6$.

Try $x = 1$: $$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$$ Yes, $x = 1$ is a root.

Now perform polynomial division: $x^3 - 6x^2 + 11x - 6$ divided by $x - 1$:

Using synthetic division with $1$:

$$\begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array}$$ So the quotient is $x^2 - 5x + 6$.

Factor $x^2 - 5x + 6$: $$(x - 2)(x - 3) = 0$$ Roots: $x = 2$ and $x = 3$.

Thus, integer solutions: $x = 1, \; x = 2, \; x = 3$.

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Step 2 – Sum of cubes of the roots

The roots are $r_1 = 1, r_2 = 2, r_3 = 3$.

We want $r_1^3 + r_2^3 + r_3^3$:

$$1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$$ Numerically, it's already $36$.

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Step 3 – Prove algebraically using Vieta's formulas

For $x^3 - 6x^2 + 11x - 6 = 0$, sum of roots $S_1 = r_1 + r_2 + r_3 = 6$.

Sum of roots taken two at a time: $S_2 = r_1 r_2 + r_2 r_3 + r_3 r_1 = 11$.

Product of roots: $S_3 = r_1 r_2 r_3 = 6$.

We know the identity: $$r_1^3 + r_2^3 + r_3^3 = S_1^3 - 3 S_1 S_2 + 3 S_3$$ Let’s verify the identity first (standard identity for sum of cubes of roots in terms of elementary symmetric sums).

Actually, known formula: $$r_1^3 + r_2^3 + r_3^3 = S_1^3 - 3 S_1 S_2 + 3 S_3$$ Where $S_1 = \sum r_i$, $S_2 = \sum_{i<j} r_i r_j$, $S_3 = r_1 r_2 r_3$.

Check: From $r_i^3 = r_i \cdot r_i^2$, but better:
Use $\sum r_i^3 = (\sum r_i)^3 - 3\sum r_i \sum_{i<j} r_i r_j + 3 r_1 r_2 r_3$ (since $r_1^2 r_2 + r_1 r_2^2 = r_1 r_2 (r_1 + r_2)$ etc., sum over all symmetric).

Derivation (known identity): $$(r_1 + r_2 + r_3)^3 = r_1^3 + r_2^3 + r_3^3 + 3(r_1 + r_2 + r_3)(r_1 r_2 + r_2 r_3 + r_3 r_1) - 3 r_1 r_2 r_3$$ Rearrange: $$r_1^3 + r_2^3 + r_3^3 = S_1^3 - 3 S_1 S_2 + 3 S_3$$

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Step 4 – Plug values

Here: $$S_1 = 6, \quad S_2 = 11, \quad S_3 = 6$$ $$r_1^3 + r_2^3 + r_3^3 = 6^3 - 3 \cdot 6 \cdot 11 + 3 \cdot 6$$ $$= 216 - 198 + 18 = 36$$

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Step 5 – Conclusion

We found integer solutions $1, 2, 3$, and using Vieta’s formulas we proved the sum of their cubes is $36$.

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$$\boxed{36}$$

Here is the step-by-step solution to find the integer roots and prove the sum of their cubes.

Part 1: Finding All Integer Solutions

1. Apply the Rational Root Theorem For the polynomial $P(x) = x^3 - 6x^2 + 11x - 6 = 0$, any integer root must be a divisor of the constant term, $-6$.

• Possible integer roots: $\pm 1, \pm 2, \pm 3, \pm 6$.

2. Test Candidates Substitute the potential roots into the equation to find a zero value.

• Test $x = 1$: $$(1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$$ Since $P(1) = 0$, $x = 1$ is a root. Consequently, $(x - 1)$ is a factor.

3. Factor the Polynomial Divide the original polynomial by $(x - 1)$ to find the remaining quadratic factor.

• Polynomial Division: $(x^3 - 6x^2 + 11x - 6) \div (x - 1) = x^2 - 5x + 6$
• Solve the Quadratic: Factor $x^2 - 5x + 6 = 0$. We need two numbers that multiply to $6$ and add to $-5$. These are $-2$ and $-3$. $$(x - 2)(x - 3) = 0$$ This yields roots $x = 2$ and $x = 3$.

Conclusion for Part 1 The set of all integer solutions is ${1, 2, 3}$.

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Part 2: Proving the Sum of Cubes Equals 36

Let the roots found in Part 1 be $x_1 = 1$, $x_2 = 2$, and $x_3 = 3$. We must calculate the sum of their cubes: $$S = x_1^3 + x_2^3 + x_3^3$$

Step 1: Cube each root

• $x_1^3 = 1^3 = 1$
• $x_2^3 = 2^3 = 8$
• $x_3^3 = 3^3 = 27$

Step 2: Sum the results $$S = 1 + 8 + 27$$ $$S = 9 + 27$$ $$S = 36$$

Final Conclusion The sum of the cubes of the roots is $36$. Q.E.D.