Let's break down the given equation step by step:

$\frac{1}{4 - X} - \frac{3}{X^2 - 4 \cdot X} = \frac{1 + X}{8 - 2 \cdot X}$

Step 1: Simplify the denominator of the second fraction

The term in the second fraction on the left-hand side is:

$X^2 - 4 \cdot X$

This can be factored as:

$X^2 - 4X = X(X - 4)$

Thus, the left-hand side becomes:

$\frac{1}{4 - X} - \frac{3}{X(X - 4)}$

Step 2: Common denominator on the left-hand side

The denominators of the two fractions on the left-hand side are $4 - X$ and $X(X - 4)$. To find a common denominator, notice that $4 - X = -(X - 4)$. This means we can rewrite the first fraction with a denominator of $X(X - 4)$.

$\frac{1}{4 - X} = -\frac{1}{X - 4}$

Now, both fractions have the same denominator:

$-\frac{1}{X - 4} - \frac{3}{X(X - 4)} = \frac{-(X + 3)}{X(X - 4)}$

So the left-hand side becomes:

$\frac{-(X + 3)}{X(X - 4)}$

Step 3: Simplify the right-hand side

On the right-hand side, we have:

$\frac{1 + X}{8 - 2 \cdot X}$

Factor the denominator on the right-hand side:

$8 - 2 \cdot X = 2(4 - X)$

So the right-hand side becomes:

$\frac{1 + X}{2(4 - X)}$

Step 4: Set the two sides equal to each other

We now have:

$\frac{-(X + 3)}{X(X - 4)} = \frac{1 + X}{2(4 - X)}$

Step 5: Cross-multiply to solve

Cross-multiplying gives:

$-(X + 3) \cdot 2(4 - X) = (1 + X) \cdot X(X - 4)$

Simplify both sides:

On the left-hand side:

$-2(X + 3)(4 - X)$

Expand:

[ -2[(X + 3)(4 - X)] = -2[(4X - X^2 + 12 - 3X)] = -2(X + 12 - X^2 - 3X) ] This becomes:

$= -2(12 - 2X - X^2) = -24 + 4X + 2X^2$

On the right-hand side:

$(1 + X)X(X - 4)$

Expand:

$= X(X^2 - 4X) + 1(X^2 - 4X)$

This simplifies to:

$X^3 - 4X^2 + X^2 - 4X = X^3 - 3X^2 - 4X$

Step 6: Solve the resulting equation

Now we equate the two sides:

$-24 + 4X + 2X^2 = X^3 - 3X^2 - 4X$

Move all terms to one side of the equation:

$0 = X^3 - 5X^2 - 8X + 24$

This is a cubic equation:

$X^3 - 5X^2 - 8X + 24 = 0$

Step 7: Solve the cubic equation

To solve this cubic equation, we can try factoring by grouping or using the Rational Root Theorem to find potential solutions.

Do you want to continue solving this cubic equation, or would you like to use an approximation method or graphing approach to find the roots?

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Relative Questions:

1. What is the Rational Root Theorem, and how can it be applied to cubic equations?
2. Can synthetic division help in solving cubic equations?
3. What is the geometric interpretation of solving cubic equations?
4. How are cubic equations related to real-world problems?
5. What numerical methods can be used to solve cubic equations efficiently?

Tip:

When solving higher-degree polynomials, always check for common factors or rational roots first to simplify the process.