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The given expression is:

\[
\frac{2x^2 y z^3}{(xyz^2)^3 + (x^2 yz)^2}
\]

Let’s simplify it step by step.

### Step 1: Expand the denominator

- The first term in the denominator is \((xyz^2)^3\):
  \[
  (xyz^2)^3 = x^3 y^3 (z^2)^3 = x^3 y^3 z^6
  \]

- The second term in the denominator is \((x^2 yz)^2\):
  \[
  (x^2 yz)^2 = (x^2)^2 (yz)^2 = x^4 y^2 z^2
  \]

### Step 2: Substitute these into the expression

Now the expression becomes:

\[
\frac{2x^2 y z^3}{x^3 y^3 z^6 + x^4 y^2 z^2}
\]

### Step 3: Factor out common terms in the denominator

In the denominator, both terms have common factors of \(x^3 y^2 z^2\), so we factor that out:

\[
x^3 y^3 z^6 + x^4 y^2 z^2 = x^3 y^2 z^2 (y z^4 + x)
\]

### Step 4: Simplify the expression

Now the expression is:

\[
\frac{2x^2 y z^3}{x^3 y^2 z^2 (y z^4 + x)}
\]

Cancel out the common factors in the numerator and the denominator:
- \(x^2\) from the numerator cancels with \(x^2\) from the denominator, leaving \(x\) in the denominator.
- \(y\) from the numerator cancels with \(y^2\) from the denominator, leaving \(y\) in the denominator.
- \(z^2\) from the numerator cancels with \(z^2\) from the denominator.

Thus, the simplified expression becomes:

\[
\frac{2}{x y (y z^4 + x)}
\]

This is the simplified form of the given expression.

Would you like further details or explanations on any steps?

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Here are five related questions to practice:
1. How would you simplify \(\frac{3x^2y^3}{(x^3 y)^2 + (x y^3)^2}\)?
2. What happens if the powers in the denominator were larger, say \( (xyz^2)^4 + (x^2yz)^3 \)?
3. How do you factor common terms in more complex expressions?
4. What is the role of exponents when expanding terms like \((xyz)^n\)?
5. How do you handle expressions involving fractional exponents?

**Tip:** Always look for common factors to simplify complex algebraic fractions.

Simplify the expression \(\frac{2x^2 y z^3}{(xyz^2)^3 + (x^2 yz)^2}\).

Learn how to simplify complex algebraic fractions like \(\frac{2x^2 y z^3}{(xyz^2)^3 + (x^2 yz)^2}\) by applying exponent rules and factoring techniques. This detailed solution is perfect for high school students studying algebra in Grades 10-12.

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