https://r2.erweima.ai/mathbot/image/1726788645144m126dd4l.jpg

Let's go through each of the problems in the image you provided:

### 1. **Calculation of \( I \)**
You are asked to calculate:
\[
I = \sqrt[3]{2^2 \cdot 2^4 \cdot 2^3 \cdot 3^2 \cdot 3^7 \cdot 2^5 \cdot 2^{40}}
\]

We can simplify this step-by-step by combining the exponents for the same base:
- For base 2: \( 2^2 \cdot 2^4 \cdot 2^3 \cdot 2^5 \cdot 2^{40} \) gives \( 2^{2+4+3+5+40} = 2^{54} \).
- For base 3: \( 3^2 \cdot 3^7 \) gives \( 3^{2+7} = 3^9 \).

Thus, \( I = \sqrt[3]{2^{54} \cdot 3^9} = 2^{18} \cdot 3^3 \).

Therefore:
\[
I = 2^{18} \cdot 3^3 = 262144 \cdot 27 = 7077888
\]

### 2. **Find \( \frac{m}{n} \) (with \( n < 0 \)) if the division gives a remainder \( R(x) = 6x - 1 \)**

The division is:
\[
\frac{24x^4 + 11x^3 + nx + m}{8x^2 + x - 3}
\]

This requires polynomial division. The remainder given is \( 6x - 1 \), and by performing the division, we can find \( n \) and \( m \) that satisfy this condition.

### 3. **Solve:**
If:
\[
x + \frac{1}{x} = 7
\]
You are asked to calculate:
\[
E = x^3 + \frac{1}{x^3} + 1
\]

This can be derived using the identity:
\[
(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})
\]
Given \( x + \frac{1}{x} = 7 \), then:
\[
7^3 = x^3 + \frac{1}{x^3} + 3 \times 7
\]
So:
\[
343 = x^3 + \frac{1}{x^3} + 21
\]
Thus:
\[
x^3 + \frac{1}{x^3} = 343 - 21 = 322
\]
Finally:
\[
E = 322 + 1 = 323
\]

### 4. **Factor the polynomial \( f(x) = x^5 - 15x^4 + 85x^3 - 225x^2 + 274x - 120 \)**

We need to factor this polynomial using binomial divisors. We can try synthetic or polynomial division, checking possible rational roots (like factors of 120 divided by factors of 1), to factor this fully.

### 5. **Reduce the expression:**
\[
E = \frac{x^2 + 11x + 24}{x^2 - 3x - 18} - \frac{x^2 + 8x + 12}{x^2 - 6x - 16}
\]

You need to factor each quadratic expression and simplify.

Would you like detailed steps for each of these problems, or do you have a specific one in mind that you'd like to focus on?

### 5 Follow-up Questions:
1. Would you like to see how the polynomial division works in Problem 2?
2. Do you need help with the factoring of quadratics in Problem 5?
3. Are you familiar with synthetic division for Problem 4, or would you prefer step-by-step long division?
4. Should I go into more detail on how to derive \( E = 323 \) in Problem 3?
5. Would you like additional examples on simplifying exponents and roots like in Problem 1?

#### Tip:
When simplifying expressions involving exponents and roots, always group terms with the same base to combine their exponents efficiently!

1. Calculate I = \sqrt[3]{2^2 \cdot 2^4 \cdot 2^3 \cdot 3^2 \cdot 3^7 \cdot 2^5 \cdot 2^{40}}. 
2. Determine \frac{m}{n} (n < 0), if the division 24x^4 + 11x^3 + nx + m by 8x^2 + x - 3 gives a remainder R(x) = 6x - 1. 
3. Solve for E if x + 1/x = 7 and E = x^3 + 1/x^3 + 1. 
4. Factor the polynomial f(x) = x^5 - 15x^4 + 85x^3 - 225x^2 + 274x - 120 using binomial divisors. 
5. Reduce E = \frac{x^2 + 11x + 24}{x^2 - 3x - 18} - \frac{x^2 + 8x + 12}{x^2 - 6x - 16}.

Explore advanced topics like polynomial division, exponentiation simplification, binomial factorization, and solving algebraic identities in this detailed mathematical problem set. Suitable for Grades 11-12.

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