We are given the equation:

$(x+1)^{-1} = x^{-1} + x$

To solve this equation, let's follow a step-by-step approach:

Step 1: Simplify the left-hand side

$(x+1)^{-1} = \frac{1}{x+1}$ So the equation becomes: $\frac{1}{x+1} = \frac{1}{x} + x$

Step 2: Combine the terms on the right-hand side

The right-hand side of the equation is: $\frac{1}{x} + x = \frac{1}{x} + \frac{x^2}{x} = \frac{1 + x^3}{x}$

Thus, the equation now becomes: $\frac{1}{x+1} = \frac{1 + x^3}{x}$

Step 3: Cross-multiply to eliminate the denominators

To get rid of the fractions, we multiply both sides by $x(x+1)$: $x \cdot 1 = (1 + x^3) \cdot (x+1)$

This simplifies to: $x = (1 + x^3)(x+1)$

Step 4: Expand the right-hand side

Expanding the right-hand side: $x = (1 + x^3)(x+1) = 1 \cdot (x+1) + x^3 \cdot (x+1)$ $x = (x+1) + x^3(x+1)$ Now expand the terms: $x = x + 1 + x^4 + x^3$ Simplify: $x = x + 1 + x^4 + x^3$

Step 5: Cancel out terms and rearrange

Cancel $x$ from both sides: $0 = 1 + x^4 + x^3$ This simplifies to: $x^4 + x^3 + 1 = 0$

Step 6: Solve the polynomial equation

Now, we have the equation: $x^4 + x^3 + 1 = 0$ This is a quartic equation, which may be challenging to solve by hand. We can attempt factoring, but in this case, the equation doesn't factor easily. Therefore, numerical methods or approximations might be needed to find the solutions.

However, a quick check reveals that there are no real solutions for this equation because the expression $x^4 + x^3 + 1$ is always positive for real values of $x$, meaning there are no real roots.

Conclusion:

There are no real solutions to the equation $(x+1)^{-1} = x^{-1} + x$.

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Would you like more details on solving quartic equations or polynomial root-finding methods?

Here are 5 related questions:

1. How do you solve higher-degree polynomial equations?
2. What methods can be used to find numerical solutions for polynomials?
3. What are the conditions for the existence of real vs. complex roots in polynomials?
4. Can you explain the general process for solving quartic equations?
5. How do we use synthetic division or other factorization techniques for complex polynomials?

Tip: When dealing with polynomial equations that don’t factor easily, numerical approximation methods, such as Newton’s method, can be very helpful!