prove that i. 5/(𝑛^5 −𝑛) and 3/(𝑛^3 −𝑛)

To prove the two given statements, we interpret them as simplifications of the expressions \( \frac{5}{n^5 - n} \) and \( \frac{3}{n^3 - n} \). The aim is to analyze and simplify each of these fractions. The key step will involve factoring the denominators, which have a common factor.

### Step 1: Factor the denominator of \( n^5 - n \)
The expression \( n^5 - n \) can be factored as follows:
\[
n^5 - n = n(n^4 - 1)
\]
Next, observe that \( n^4 - 1 \) is a difference of squares:
\[
n^4 - 1 = (n^2 - 1)(n^2 + 1)
\]
Further, \( n^2 - 1 \) is also a difference of squares:
\[
n^2 - 1 = (n - 1)(n + 1)
\]
Thus, the full factorization of \( n^5 - n \) is:
\[
n^5 - n = n(n - 1)(n + 1)(n^2 + 1)
\]
This gives us the factored denominator for the first expression.

### Step 2: Factor the denominator of \( n^3 - n \)
Similarly, the expression \( n^3 - n \) can be factored:
\[
n^3 - n = n(n^2 - 1)
\]
We already know that \( n^2 - 1 = (n - 1)(n + 1) \), so the full factorization is:
\[
n^3 - n = n(n - 1)(n + 1)
\]

### Step 3: Analyze the simplified forms
#### Expression 1: \( \frac{5}{n^5 - n} \)
Using the factorization from Step 1, the first expression becomes:
\[
\frac{5}{n(n - 1)(n + 1)(n^2 + 1)}
\]
This is the fully factored form of the denominator, and no further simplification can be made for the given numerator.

#### Expression 2: \( \frac{3}{n^3 - n} \)
Using the factorization from Step 2, the second expression becomes:
\[
\frac{3}{n(n - 1)(n + 1)}
\]
Again, this is the fully factored form of the denominator, and no further simplification can be made for the given numerator.

### Step 4: Conclusion
Both expressions have been factored, and no further simplification is possible. The expressions \( \frac{5}{n^5 - n} \) and \( \frac{3}{n^3 - n} \) cannot be simplified further unless \( n \) is assigned a specific value.

Would you like more details or have any questions?

### Related Questions:
1. How would the expressions change if \( n \) were a specific number?
2. Can the expression \( n^5 - n \) have complex roots?
3. How would these factorizations change for higher powers of \( n \)?
4. What happens to these expressions as \( n \) approaches infinity?
5. How can these expressions be used in number theory?

#### Tip:
Always check for common factors in expressions involving powers of variables; factorizations often reveal simplifications.

prove that i. 5/(n^5 −n) and 3/(n^3 −n)

Learn how to factor polynomial expressions such as 5/(n^5 − n) and 3/(n^3 − n). This problem focuses on algebraic concepts, including the factorization of polynomial terms and the difference of squares theorem. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation