Question

$\frac{4y}{x^{6a}-27}cdot\frac{x^{4a}-9}{x^{2a}y + 3y}$

Explanation:

Step1: Factor the expressions

We know that \(x^{6a}-27=(x^{2a})^3 - 3^3\). By the difference - of - cubes formula \(A^3 - B^3=(A - B)(A^2+AB + B^2)\), so \(x^{6a}-27=(x^{2a}-3)(x^{4a}+3x^{2a}+9)\). Also, \(x^{4a}-9=(x^{2a})^2 - 3^2\), and by the difference - of - squares formula \(A^2 - B^2=(A + B)(A - B)\), so \(x^{4a}-9=(x^{2a}+3)(x^{2a}-3)\). And \(x^{2a}y + 3y=y(x^{2a}+3)\).

Step2: Rewrite the product of fractions

The original expression \(\frac{4y}{x^{6a}-27}\cdot\frac{x^{4a}-9}{x^{2a}y + 3y}\) can be rewritten as \(\frac{4y}{(x^{2a}-3)(x^{4a}+3x^{2a}+9)}\cdot\frac{(x^{2a}+3)(x^{2a}-3)}{y(x^{2a}+3)}\).

Step3: Cancel out the common factors

Cancel out the common factors \((x^{2a}-3)\), \((x^{2a}+3)\) and \(y\) in the numerator and the denominator. We get \(\frac{4}{x^{4a}+3x^{2a}+9}\).

Answer:

\(\frac{4}{x^{4a}+3x^{2a}+9}\)