To simplify the expression $\left(\frac{5a^5}{4b^{1/3}}\right)^{3/2}$, let's break it down step by step.

Step 1: Apply the power of a quotient rule

The general rule for simplifying an expression raised to a power is: $\left( \frac{x}{y} \right)^n = \frac{x^n}{y^n}$ So, for our expression: $\left(\frac{5a^5}{4b^{1/3}}\right)^{3/2} = \frac{(5a^5)^{3/2}}{(4b^{1/3})^{3/2}}$

Step 2: Simplify the numerator

We apply the power rule $(x^m)^n = x^{m \cdot n}$ to simplify each part: $(5a^5)^{3/2} = 5^{3/2} \cdot (a^5)^{3/2}$ $(a^5)^{3/2} = a^{5 \cdot \frac{3}{2}} = a^{15/2}$ Thus, the numerator becomes: $5^{3/2} \cdot a^{15/2}$

Step 3: Simplify the denominator

Similarly, we simplify the denominator: $(4b^{1/3})^{3/2} = 4^{3/2} \cdot (b^{1/3})^{3/2}$ $(b^{1/3})^{3/2} = b^{(1/3) \cdot (3/2)} = b^{1/2}$ Thus, the denominator becomes: $4^{3/2} \cdot b^{1/2}$

Step 4: Final expression

Now the expression becomes: $\frac{5^{3/2} \cdot a^{15/2}}{4^{3/2} \cdot b^{1/2}}$

Step 5: Simplify the constants

We can leave $5^{3/2}$ and $4^{3/2}$ as they are or express them in radical form: $5^{3/2} = 5 \cdot \sqrt{5}, \quad 4^{3/2} = 8 \quad (\text{since } 4^{3/2} = (2^2)^{3/2} = 2^3 = 8)$ Thus, the simplified expression is: $\frac{5 \cdot \sqrt{5} \cdot a^{15/2}}{8 \cdot b^{1/2}}$

Final Answer:

$\frac{5 \cdot \sqrt{5} \cdot a^{15/2}}{8 \cdot b^{1/2}}$

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Would you like further details on any of these steps or have any other questions?

Related questions:

1. How do you simplify rational exponents with different bases?
2. What is the rule for simplifying expressions with fractional exponents?
3. How do you convert between radical and exponent notation?
4. How would you handle a rational exponent in the denominator of a fraction?
5. Can this approach be extended to more complex expressions involving multiple variables and powers?

Tip:

When simplifying expressions with rational exponents, remember that fractional exponents represent roots. The numerator of the fraction is the power, and the denominator is the root. This can be helpful in breaking down the expression.