Solving Algebraic Expression with a Given Condition

The problem asks us to find the value of a complex algebraic expression given a linear relationship between the variables \(x\) and \(a\).

The given condition is \(4x + 3a = 0\). We can use this condition to express \(x\) in terms of \(a\):

From \(4x + 3a = 0\), we subtract \(3a\) from both sides:

\(4x = -3a\)

Now, divide both sides by 4:

\(x = -\frac{3a}{4}\)

We need to evaluate the expression: \(\frac{{{x^2}\; + \;ax\; + \;{a^2}}}{{{x^3} - {a^3}}} - \;\frac{{{x^2} - \;ax\; + \;{a^2}}}{{{x^3}\; + \;{a^3}}}\)

Let's simplify the expression first before substituting the value of \(x\). We can use the factorization formulas for the difference and sum of cubes:

• Difference of cubes: \(p^3 - q^3 = (p - q)(p^2 + pq + q^2)\)
• Sum of cubes: \(p^3 + q^3 = (p + q)(p^2 - pq + q^2)\)

Applying these formulas to the denominators of the given expression:

• \(x^3 - a^3 = (x - a)(x^2 + ax + a^2)\)
• \(x^3 + a^3 = (x + a)(x^2 - ax + a^2)\)

Substitute these factored forms back into the expression:

\(\frac{{{x^2}\; + \;ax\; + \;{a^2}}}{{(x - a)(x^2 + ax + a^2)}} - \;\frac{{{x^2} - \;ax\; + \;{a^2}}{}}{(x + a)(x^2 - ax + a^2)}\)

Notice that the term \((x^2 + ax + a^2)\) appears in the numerator and denominator of the first fraction, and the term \((x^2 - ax + a^2)\) appears in the numerator and denominator of the second fraction. Assuming \(x^2 + ax + a^2 \neq 0\) and \(x^2 - ax + a^2 \neq 0\), we can cancel these terms:

\(\frac{1}{x - a} - \;\frac{1}{x + a}\)

Now, combine these two fractions by finding a common denominator, which is \((x - a)(x + a)\):

\(\frac{(x + a) - (x - a)}{(x - a)(x + a)}\)

Expand the numerator:

\(\frac{x + a - x + a}{x^2 - a^2}\)

Simplify the numerator:

\(\frac{2a}{x^2 - a^2}\)

Now we substitute the value of \(x\) we found from the given condition: \(x = -\frac{3a}{4}\)

\(\frac{2a}{\left(-\frac{3a}{4}\right)^2 - a^2}\)

Calculate the square in the denominator:

\(\left(-\frac{3a}{4}\right)^2 = \left(\frac{-3a}{4}\right) \times \left(\frac{-3a}{4}\right) = \frac{(-3a)(-3a)}{4 \times 4} = \frac{9a^2}{16}\)

Substitute this back into the denominator:

\(\frac{2a}{\frac{9a^2}{16} - a^2}\)

Find a common denominator for the terms in the denominator (\(16\)):

\(\frac{2a}{\frac{9a^2}{16} - \frac{16a^2}{16}}\)

\(\frac{2a}{\frac{9a^2 - 16a^2}{16}}\)

\(\frac{2a}{\frac{-7a^2}{16}}\)

To divide by a fraction, multiply by its reciprocal:

\(2a \times \frac{16}{-7a^2}\)

\(\frac{2a \times 16}{-7a^2}\)

\(\frac{32a}{-7a^2}\)

We can cancel \(a\) from the numerator and denominator, assuming \(a \neq 0\):

\(\frac{32}{-7a}\)

This can be written as \(-\frac{32}{7a}\).

This matches one of the provided options.

Revision Table: Key Concepts Reviewed

Additional Information: Algebraic Identities

Understanding algebraic identities is crucial for simplifying expressions quickly and accurately. Here are some common identities:

• \((p+q)^2 = p^2 + 2pq + q^2\)
• \((p-q)^2 = p^2 - 2pq + q^2\)
• \((p+q)(p-q) = p^2 - q^2\)
• \((p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3\)
• \((p-q)^3 = p^3 - 3p^2q + 3pq^2 - q^3\)
• \(p^3 + q^3 = (p+q)(p^2 - pq + q^2)\)
• \(p^3 - q^3 = (p-q)(p^2 + pq + q^2)\)

In this problem, we specifically used the difference and sum of cubes identities for factorization, and the difference of squares identity in the common denominator \((x-a)(x+a) = x^2 - a^2\).

It's important to note the conditions under which these simplifications are valid. When we cancelled terms like \((x^2 + ax + a^2)\), we assumed they are non-zero. Also, the final step of cancelling \(a\) assumes \(a \neq 0\). If \(a=0\), the original condition \(4x+3a=0\) implies \(4x=0\), so \(x=0\). In this case, the original expression's denominators would be \(0^3 - 0^3 = 0\) and \(0^3 + 0^3 = 0\), making the expression undefined. Therefore, we are working under the assumption that \(a \neq 0\).