(d) $$ \left(p^{-\frac{1}{3}} q^{2} ight)^{5} imes \sqrt[3]{27 p^{-3} q^{2}} $$

Question

(d) $$ \left(p^{-\frac{1}{3}} q^{2}ight)^{5} 	imes \sqrt[3]{27 p^{-3} q^{2}} $$

UpStudy AI · Accepted Answer

Let us simplify the expression 
$$
\left(p^{-\frac{1}{3}}q^2\right)^5 \times \sqrt[3]{27p^{-3}q^2}.
$$

**Step 1. Simplify $$\left(p^{-\frac{1}{3}}q^2\right)^5$$:**

Using the exponent rule $$(a^m)^n = a^{mn}$$, we have:
$$
\left(p^{-\frac{1}{3}}q^2\right)^5 = p^{-\frac{1}{3}\times 5} \, q^{2\times 5} = p^{-\frac{5}{3}}q^{10}.
$$

**Step 2. Simplify $$\sqrt[3]{27p^{-3}q^2}$$:**

Recall that the cube root can be written with an exponent $$\frac{1}{3}$$:
$$
\sqrt[3]{27p^{-3}q^2} = \left(27p^{-3}q^2\right)^{\frac{1}{3}}.
$$

Now, apply the exponent rule to each factor:
- $$27^{\frac{1}{3}} = 3$$ (since $$27=3^3$$),
- $$\left(p^{-3}\right)^{\frac{1}{3}} = p^{-1}$$,
- $$\left(q^2\right)^{\frac{1}{3}} = q^{\frac{2}{3}}$$.

Thus,
$$
\sqrt[3]{27p^{-3}q^2} = 3p^{-1}q^{\frac{2}{3}}.
$$

**Step 3. Multiply the simplified parts:**

Now, multiply the two results:
$$
p^{-\frac{5}{3}}q^{10} \times 3p^{-1}q^{\frac{2}{3}}.
$$

First, combine the coefficients and then the like bases:
- For $$p$$:
  $$
  p^{-\frac{5}{3}} \times p^{-1} = p^{-\frac{5}{3} - 1} = p^{-\frac{5}{3} - \frac{3}{3}} = p^{-\frac{8}{3}}.
  $$
- For $$q$$:
  $$
  q^{10} \times q^{\frac{2}{3}} = q^{10 + \frac{2}{3}} = q^{\frac{30}{3} + \frac{2}{3}} = q^{\frac{32}{3}}.
  $$

Including the coefficient 3, we obtain:
$$
3p^{-\frac{8}{3}}q^{\frac{32}{3}}.
$$

Thus, the simplified expression is:
$$
\boxed{3p^{-\frac{8}{3}}q^{\frac{32}{3}}}.
$$
The simplified expression is $$3p^{-\frac{8}{3}}q^{\frac{32}{3}}$$.

(d)

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