Solve the Rational Expression: Find x in \(\frac{x^3+1}{(x+1)^2}=x\)

To solve the equation $\frac{x^3 + 1}{(x+1)^2} = x$, we will follow these steps:

• Step 1: Set up the equation for solving by cross-multiplying.
• Step 2: Simplify and solve the resulting polynomial equation.
• Step 3: Solve for the values of $x$.

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

$(x^3 + 1) = x \cdot (x+1)^2$

Expand the right-hand side:

$x \cdot (x^2 + 2x + 1) = x^3 + 2x^2 + x$

Step 2: Set the expanded equation equal:

$x^3 + 1 = x^3 + 2x^2 + x$

Cancel $x^3$ from both sides:

$1 = 2x^2 + x$

Re-arrange the equation to form a standard quadratic equation:

$0 = 2x^2 + x - 1$

Step 3: Solve the quadratic equation using the quadratic formula:

$\text{Here, } a = 2, \, b = 1, \, c = -1.$

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2}$

Calculate the discriminant and simplify:

$x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4}$

Simplify further:

$x = \frac{-1 \pm 3}{4}$

This gives the solutions:

$x = \frac{2}{4} = \frac{1}{2}$ $x = \frac{-4}{4} = -1$

Since $x = -1$ would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is $x = \frac{1}{2}$.