https://file.aiquickdraw.com/mathbot/file/1728718926019qrfvng2t.png

Let's break this down step-by-step:

### 1. Solve for \(x\) from the equation:
\[
x^2 + \frac{9}{x^2} = 10
\]

We multiply both sides by \(x^2\) to clear the fraction:
\[
x^4 + 9 = 10x^2
\]

Rearrange to form a quadratic equation in terms of \(u = x^2\):
\[
x^4 - 10x^2 + 9 = 0
\]

Now, substitute \(u = x^2\), so the equation becomes:
\[
u^2 - 10u + 9 = 0
\]

Solve this quadratic using the quadratic formula:
\[
u = \frac{10 \pm \sqrt{10^2 - 4 \cdot 9}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}
\]

So:
\[
u = \frac{10 + 8}{2} = 9 \quad \text{and} \quad u = \frac{10 - 8}{2} = 1
\]

Thus:
\[
x^2 = 9 \quad \text{or} \quad x^2 = 1
\]

Therefore:
\[
x = \pm 3 \quad \text{or} \quad x = \pm 1
\]

### 2. Apply the inequality condition:
We are given that \(x^3 - 3x > 0\). Let’s first factor the expression:
\[
x^3 - 3x = x(x - \sqrt{3})(x + \sqrt{3})
\]

We need the product to be positive. This depends on the sign of each factor across different ranges of \(x\). We solve for:

- **Sign analysis of \(x(x - \sqrt{3})(x + \sqrt{3})\)**:
  - **For \(x > \sqrt{3}\)**: All factors are positive.
  - **For \(-\sqrt{3} < x < \sqrt{3}\)**: Signs alternate; the product is negative.
  - **For \(x < -\sqrt{3}\)**: All factors are negative, so the product is negative.

Thus, the inequality \(x^3 - 3x > 0\) holds **only when \(x > \sqrt{3}\)** or **\(x < -\sqrt{3}\)**.

### 3. Combining the Solutions:
From the equation \(x^2 + \frac{9}{x^2} = 10\), the possible solutions are \(x = \pm 3\) and \(x = \pm 1\). Now, we need to keep only those solutions that satisfy the inequality \(x^3 - 3x > 0\).

- **For \(x = 3\)**: \(3^3 - 3(3) = 27 - 9 = 18 > 0\).
- **For \(x = -3\)**: \((-3)^3 - 3(-3) = -27 + 9 = -18 < 0\).
- **For \(x = \pm 1\)**: \(1^3 - 3(1) = -2 < 0\), \((-1)^3 - 3(-1) = 2 > 0\), but the sign is opposite to the positive requirement.

Thus, the only valid solution is:
\[
x = 3
\]

### 4. Show that \(y = x^3 - 3x\) is increasing at \(x = 1\):
We need to compute the derivative of \(y = x^3 - 3x\) to check if it is increasing at \(x = 1\).

\[
\frac{dy}{dx} = 3x^2 - 3
\]

Evaluate the derivative at \(x = 1\):
\[
\frac{dy}{dx} \Big|_{x = 1} = 3(1)^2 - 3 = 3 - 3 = 0
\]

Since the derivative is **0** at \(x = 1\), the function is neither increasing nor decreasing at this point (it is a critical point).

---

### Summary:
- The solution to the equation \(x^2 + \frac{9}{x^2} = 10\) that also satisfies the inequality \(x^3 - 3x > 0\) is:
  \[
  x = 3
  \]
- The derivative of \(y = x^3 - 3x\) at \(x = 1\) is 0, so the function is neither increasing nor decreasing at \(x = 1\).

---

Do you have any questions or need further details?

### Related Questions:
1. How do we determine intervals of increase and decrease for a function?
2. Can a function have multiple solutions that partially satisfy a given inequality?
3. What is the significance of critical points in calculus?
4. How does the quadratic formula generalize to higher-degree polynomials?
5. How can we visualize inequalities graphically?

### Tip:
When analyzing inequalities involving polynomials, sign charts can be a useful tool to identify solution intervals accurately.

Consider the equation x^2 + 9/x^2 = 10. Solve for x such that the inequality x^3 - 3x > 0 is also satisfied. Additionally, show that y = x^3 - 3x is increasing at x = 1.

This problem involves solving the equation x^2 + 9/x^2 = 10 and finding the valid solutions that satisfy the inequality x^3 - 3x > 0. Additionally, we check whether y = x^3 - 3x is increasing at x = 1 by finding its derivative. Ideal for students in Grades 10-12.

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