simplify step by step \frac{\left(2x^2\cdot y\right)^2\cdot 5x\cdot y}{\left(3x\cdot y^3\right)^3}

Let's simplify the expression step by step:

\[
\frac{\left( (2x^2 \cdot y)^2 \cdot 5x \cdot y \right)}{\left(3x \cdot y^3\right)^3}
\]

### Step 1: Expand the powers in the numerator
We first expand \((2x^2 \cdot y)^2\). Using the power of a product rule, \((ab)^2 = a^2 \cdot b^2\), we get:
\[
(2x^2 \cdot y)^2 = 2^2 \cdot (x^2)^2 \cdot y^2 = 4x^4y^2
\]

Now, substitute this back into the numerator:
\[
\text{Numerator} = (4x^4y^2) \cdot 5x \cdot y = 4 \cdot 5 \cdot x^4 \cdot x \cdot y^2 \cdot y = 20x^5y^3
\]

So the numerator becomes:
\[
20x^5y^3
\]

### Step 2: Expand the denominator
Now expand the denominator \((3x \cdot y^3)^3\). Again, use the power of a product rule:
\[
(3x \cdot y^3)^3 = 3^3 \cdot x^3 \cdot (y^3)^3 = 27x^3y^9
\]

So the denominator becomes:
\[
27x^3y^9
\]

### Step 3: Simplify the fraction
Now, simplify the expression:
\[
\frac{20x^5y^3}{27x^3y^9}
\]

For the \(x\)-terms:
\[
\frac{x^5}{x^3} = x^{5-3} = x^2
\]

For the \(y\)-terms:
\[
\frac{y^3}{y^9} = y^{3-9} = y^{-6} = \frac{1}{y^6}
\]

### Final expression
Now combine everything:
\[
\frac{20x^5y^3}{27x^3y^9} = \frac{20x^2}{27y^6}
\]

Thus, the simplified expression is:
\[
\boxed{\frac{20x^2}{27y^6}}
\]

Would you like further details or clarifications?

Here are 5 questions to explore:

1. How do you simplify expressions with powers in both the numerator and denominator?
2. What is the power of a product rule, and how is it applied?
3. How does exponent subtraction work when simplifying variables?
4. How do negative exponents impact the final expression?
5. What would happen if the denominator had a higher degree for \(x\)?

**Tip:** Always apply exponent rules carefully when simplifying variables.

Learn how to simplify the expression \frac{(2x^2\cdot y)^2\cdot 5x\cdot y}{(3x\cdot y^3)^3} using exponent rules. This step-by-step guide covers key algebraic concepts such as the power of a product rule, exponent subtraction, and negative exponents. Ideal for students in Grades 9-12.

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