Just solve x by changing the subject of the formula

To solve for \( x \) by changing the subject of the formula, we'll simplify and manipulate the given equation step by step:

\[ \frac{b}{x - a} = \frac{bx}{x^2 + ax + a^2} - \frac{b(2a - a^2)}{x^3 - a^3} \]

Here are the steps:

1. **Combine the terms on the right-hand side:**

First, note that \( x^3 - a^3 = (x - a)(x^2 + ax + a^2) \). 

\[ \frac{b}{x - a} = \frac{bx}{x^2 + ax + a^2} - \frac{b(2a - a^2)}{(x - a)(x^2 + ax + a^2)} \]

2. **Combine the fractions on the right-hand side:**

We can rewrite the right-hand side with a common denominator:

\[ \frac{b}{x - a} = \frac{bx(x^2 + ax + a^2) - b(2a - a^2)}{(x^2 + ax + a^2)(x - a)} \]

Simplify the numerator:

\[ bx(x^2 + ax + a^2) - b(2a - a^2) \]
\[ = bx^3 + bax^2 + ba^2x - 2ab + ba^2 \]

3. **Equate the numerators and denominators:**

Since the denominators are equal, we can equate the numerators directly:

\[ b = bx(x^2 + ax + a^2) - b(2a - a^2) \]

\[ b(x - a)(x^2 + ax + a^2) = bx(x^2 + ax + a^2) - b(2a - a^2) \]

Let's rewrite this equation without the denominators:

\[ b(x - a) = bx - b(2a - a^2) \]

4. **Expand and simplify the equation:**

\[ b(x - a) = bx - b(2a - a^2) \]
\[ b(x - a) = bx - b2a + ba^2 \]

Notice that the terms involving \( x \) on both sides are similar. Let's solve for \( x \) explicitly.

\[ b(x - a) = bx - b2a + ba^2 \]

Subtract \( bx \) from both sides:

\[ b(x - a - x) = -b2a + ba^2 \]

\[ -ba = -b2a + ba^2 \]

Divide by \( b \):

\[ -a = -2a + a^2 \]

Rearrange and solve the quadratic equation:

\[ a^2 - a - 2a = 0 \]
\[ a^2 - 3a = 0 \]
\[ a(a - 3) = 0 \]

So, \( a = 0 \) or \( a = 3 \).

However, if \( a = 0 \):

\[ \frac{b}{x - 0} = \frac{bx}{x^2} - \frac{b(2(0) - 0^2)}{x^3 - 0^3} \]
\[ \frac{b}{x} = \frac{bx}{x^2} \]
\[ b = b \]

This case doesn't help us find \( x \).

If \( a = 3 \):

\[ \frac{b}{x - 3} = \frac{bx}{x^2 + 3x + 9} - \frac{b(2(3) - 3^2)}{x^3 - 27} \]

This is more complex, and we'll need to evaluate this specifically for a solution to \( x \). 

Given the setup and the complexity of the equation, we typically use numerical methods or more advanced algebraic manipulation tools to find exact \( x \) values for specific \( a \) and \( b \).

Would you like me to proceed with a specific value of \( a \) or \( b \), or use a numerical method to approximate \( x \)?

**Related Questions:**

1. How to solve cubic equations numerically?
2. What are the standard methods for solving higher-order polynomials?
3. How does factoring assist in solving polynomial equations?
4. What role do discriminants play in solving polynomial equations?
5. How to simplify complex algebraic fractions?
6. Can software tools aid in solving complex algebraic equations?
7. How does the Rational Root Theorem apply to solving polynomial equations?
8. What is synthetic division, and how can it be used to solve polynomial equations?

**Tip:** When dealing with complex equations, breaking them into simpler parts and solving step-by-step can make the process more manageable and easier to understand.

Learn how to solve for x by changing the subject of the formula in a complex equation involving fractions and manipulation steps. This detailed guide covers algebraic concepts, quadratic equations, and advanced fraction manipulation. Suitable for advanced high school students and above.

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