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Let's break down the problem in two parts:

### Part 1: Solving the equation \( x^2 + \frac{9}{x^2} = 10 \)
We need to solve for \( x \) such that the inequality \( x^3 - 3x > 0 \) is also satisfied.

#### Step 1: Solve the equation \( x^2 + \frac{9}{x^2} = 10 \)
Multiply both sides by \( x^2 \) to eliminate the fraction:

\[
x^4 + 9 = 10x^2
\]

Rearranging terms gives:

\[
x^4 - 10x^2 + 9 = 0
\]

Let \( z = x^2 \), so the equation becomes a quadratic in \( z \):

\[
z^2 - 10z + 9 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
z = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}
\]

\[
z = \frac{10 \pm \sqrt{100 - 36}}{2}
\]

\[
z = \frac{10 \pm \sqrt{64}}{2}
\]

\[
z = \frac{10 \pm 8}{2}
\]

Thus, \( z = 9 \) or \( z = 1 \).

Since \( z = x^2 \), we have two cases:

1. \( x^2 = 9 \) → \( x = \pm 3 \)
2. \( x^2 = 1 \) → \( x = \pm 1 \)

So, \( x = -3, -1, 1, 3 \).

#### Step 2: Satisfy the inequality \( x^3 - 3x > 0 \)
Now we need to check which of these solutions satisfy \( x^3 - 3x > 0 \).

Factor the inequality:

\[
x^3 - 3x = x(x^2 - 3)
\]

This implies:

\[
x(x^2 - 3) > 0
\]

The solutions are determined by analyzing the sign changes:

- \( x(x^2 - 3) = 0 \) when \( x = 0 \) or \( x = \pm \sqrt{3} \).
- Test intervals based on these roots: \( (-\infty, -\sqrt{3}), (-\sqrt{3}, 0), (0, \sqrt{3}), (\sqrt{3}, \infty) \).

In the interval \( (0, \sqrt{3}) \), the inequality is negative. In the intervals \( (-\infty, -\sqrt{3}) \) and \( (\sqrt{3}, \infty) \), it is positive.

Now, check which values from our solutions for \( x \) satisfy this:

- \( x = 3 \): \( 3(3^2 - 3) = 3(9 - 3) = 18 > 0 \) (satisfies inequality).
- \( x = -3 \): \( -3((-3)^2 - 3) = -3(9 - 3) = -18 < 0 \) (does not satisfy inequality).
- \( x = 1 \): \( 1(1^2 - 3) = 1(1 - 3) = -2 < 0 \) (does not satisfy inequality).
- \( x = -1 \): \( -1((-1)^2 - 3) = -1(1 - 3) = 2 > 0 \) (satisfies inequality).

Thus, \( x = 3 \) and \( x = -1 \) satisfy both the equation and the inequality.

### Part 2: Show that \( y = x^3 - 3x \) is increasing at \( x = 1 \)
To determine whether \( y = x^3 - 3x \) is increasing at \( x = 1 \), we need to check the derivative \( y' \).

#### Step 1: Find the derivative
Differentiate \( y = x^3 - 3x \):

\[
y' = 3x^2 - 3
\]

#### Step 2: Evaluate the derivative at \( x = 1 \)

\[
y'(1) = 3(1)^2 - 3 = 3 - 3 = 0
\]

Since \( y'(1) = 0 \), the function is neither increasing nor decreasing at \( x = 1 \). Thus, at \( x = 1 \), the function has a critical point, possibly a local minimum or maximum.

---

Let me know if you need more details or further clarification on any part of the solution!

### 5 Related Questions:
1. How do you find critical points of a function based on its derivative?
2. Can you graph \( y = x^3 - 3x \) and identify where it increases or decreases?
3. What role do the roots of the equation \( x^2 + \frac{9}{x^2} = 10 \) play in determining the behavior of the function?
4. How do you test intervals to solve inequalities involving cubic functions?
5. Why does the derivative being zero at a point suggest a possible extremum?

### Tip:
When solving inequalities, factor the expression to identify critical points and use a sign chart to test different intervals for sign changes.

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.

Learn how to solve the equation x^2 + 9/x^2 = 10 and determine which values satisfy the inequality x^3 - 3x > 0. Additionally, explore how to show that y = x^3 - 3x is increasing at x = 1 using derivatives. This problem covers advanced algebra and calculus concepts, suitable for high school students.

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