3005^2 ( 3004^2 - 3003) / (3004^2 - 1) ( 3004^3+ 1) x 3003^2 (3004^2 + 3005) / 3004^3 - 1

The expression you've provided is complex, and I will work through simplifying it step by step. Here's the expression:

\[
\frac{{3005^2 \cdot (3004^2 - 3003)}}{{(3004^2 - 1) \cdot (3004^3 + 1)}} \times \frac{{3003^2 \cdot (3004^2 + 3005)}}{{3004^3 - 1}}
\]

Let's break it down into manageable parts:

### Numerator:
\[
3005^2 \cdot (3004^2 - 3003) \times 3003^2 \cdot (3004^2 + 3005)
\]

### Denominator:
\[
(3004^2 - 1) \cdot (3004^3 + 1) \cdot (3004^3 - 1)
\]

The denominator has factorizations that we can simplify:
- \( 3004^2 - 1 \) is a difference of squares: 
  \[
  3004^2 - 1 = (3004 - 1)(3004 + 1) = 3003 \cdot 3005
  \]
- \( 3004^3 + 1 \) is a sum of cubes, which can be factored as:
  \[
  3004^3 + 1 = (3004 + 1)(3004^2 - 3004 + 1)
  \]
- \( 3004^3 - 1 \) is a difference of cubes, which can be factored as:
  \[
  3004^3 - 1 = (3004 - 1)(3004^2 + 3004 + 1) = 3003(3004^2 + 3004 + 1)
  \]

Substituting these into the original expression will allow us to simplify it.

Let me compute this for you.After simplifying the given expression, the result is \( 1.0 \).

Would you like further details on the steps or any clarifications?

Here are five related questions to expand your understanding:
1. How can we use factorization techniques like difference of squares or cubes in other problems?
2. What are the common patterns for factoring polynomials?
3. How can large numbers be simplified efficiently in mathematical expressions?
4. What are some real-life applications of simplifying complex algebraic expressions?
5. How does simplification help when solving equations involving multiple terms?

**Tip**: Whenever dealing with large expressions, look for factorization opportunities like difference of squares, cubes, or sum of cubes to make the problem more manageable!

This step-by-step guide simplifies a complex algebraic expression involving squares and cubes. Learn key factorization techniques like the difference of squares, sum of cubes, and difference of cubes to solve the problem. Ideal for students in Grades 10-12.

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