DeepSeek vs Perplexity for Math

Alright, let's go step-by-step.

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Step 1: Solve the cubic equation for integer solutions.

We have:

$$x^3 - 6x^2 + 11x - 6 = 0$$

Possible integer roots are divisors of the constant term $-6$:

$$\pm 1, \pm 2, \pm 3, \pm 6$$

Test $x = 1$:

$$1 - 6 + 11 - 6 = 0 \quad \checkmark$$

So $x = 1$ is a root.

Divide by $(x-1)$:

$$x^3 - 6x^2 + 11x - 6 \div (x-1)$$

Using synthetic division: Coefficients: $1, -6, 11, -6$

Bring down $1$, multiply by $1$: $1 \times 1 = 1$, add to $-6$ → $-5$
Multiply: $1 \times (-5) = -5$, add to $11$ → $6$
Multiply: $1 \times 6 = 6$, add to $-6$ → $0$

Quotient: $x^2 - 5x + 6$.

Factor:

$$x^2 - 5x + 6 = (x - 2)(x - 3)$$

So the roots are:

$$x = 1, \quad x = 2, \quad x = 3$$

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Step 2: Let’s verify the roots satisfy the equation.

For $x=1$: $1 - 6 + 11 - 6 = 0$ ✔
For $x=2$: $8 - 24 + 22 - 6 = 0$ ✔
For $x=3$: $27 - 54 + 33 - 6 = 0$ ✔

All integer roots: $\{1, 2, 3\}$.

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Step 3: Prove that the sum of cubes of the roots equals 36.

Let the roots be $r_1 = 1$, $r_2 = 2$, $r_3 = 3$.

$$r_1^3 + r_2^3 + r_3^3 = 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$$

That’s it — the sum is clearly $36$.

But let's show this using Vieta's formulas for practice:

From $x^3 - 6x^2 + 11x - 6 = 0$:

Let $p = r_1 + r_2 + r_3 = 6$
Let $q = r_1 r_2 + r_2 r_3 + r_3 r_1 = 11$
Let $r = r_1 r_2 r_3 = 6$

We recall the identity:

$$r_1^3 + r_2^3 + r_3^3 = (r_1 + r_2 + r_3)^3 - 3(r_1 + r_2 + r_3)(r_1 r_2 + r_2 r_3 + r_3 r_1) + 3 r_1 r_2 r_3$$

Substitute:

$$r_1^3 + r_2^3 + r_3^3 = 6^3 - 3(6)(11) + 3 \cdot 6$$

$$= 216 - 198 + 18$$

$$= 36$$

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Final Answer: $$\boxed{36}$$

The integer solutions are $x = 1, 2, 3$, and the sum of their cubes is $36$.

The integer solutions are $x = 1$, $x = 2$, and $x = 3$. The sum of the cubes of the roots is $1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$.

Finding the integer solutions step-by-step

The equation is $x^3 - 6x^2 + 11x - 6 = 0$.

Possible rational roots are factors of 6 divided by factors of 1, so $\pm1, \pm2, \pm3, \pm6$[1].

Test $x=1$:
$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. So $x=1$ is a root.

Factor: $(x-1)(x^2 - 5x + 6) = 0$.
$x^2 - 5x + 6 = (x-2)(x-3)$.

Roots: $x=1, 2, 3$.

Proving the sum of cubes equals 36

Let roots be $\alpha=1$, $\beta=2$, $\gamma=3$.

For $x^3 + ax^2 + bx + c = 0$, sum of cubes is $\alpha^3 + \beta^3 + \gamma^3 = -a(\alpha^2 + \beta^2 + \gamma^2) - b(\alpha + \beta + \gamma) - 3c$[5].

Here, $a=-6$, $b=11$, $c=-6$.

By Vieta:
$\alpha + \beta + \gamma = 6$,
$\alpha\beta + \alpha\gamma + \beta\gamma = 11$,
$\alpha\beta\gamma = 6$.

Then $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 6^2 - 2(11) = 36 - 22 = 14$.

Sum of cubes: $-(-6)(14) - 11(6) - 3(-6) = 6 \cdot 14 - 66 + 18 = 84 - 66 + 18 = 36$[4][5].