Practice sheet featuring six problems focused on simplifying complex radical expressions containing variables and exponents.

Here are the step-by-step solutions for each problem on the worksheet.

1) Simplify: $3\sqrt[5]{s^4} + 16s \cdot s^{-\frac{1}{5}} - 5\sqrt[5]{s^{11}} + 10s \cdot s^{-\frac{1}{5}}$

Step 1: Convert all radical terms to exponential form using the rule $\sqrt[n]{x^m} = x^{\frac{m}{n}}$.
* $3\sqrt[5]{s^4}$ becomes $3s^{\frac{4}{5}}$
* $-5\sqrt[5]{s^{11}}$ becomes $-5s^{\frac{11}{5}}$

The expression is now:
$3s^{\frac{4}{5}} + 16s \cdot s^{-\frac{1}{5}} - 5s^{\frac{11}{5}} + 10s \cdot s^{-\frac{1}{5}}$

Step 2: Simplify the terms with multiplication using the product rule $x^a \cdot x^b = x^{a+b}$. Remember that $s = s^1$.
* $16s^1 \cdot s^{-\frac{1}{5}} = 16s^{1 - \frac{1}{5}} = 16s^{\frac{5}{5} - \frac{1}{5}} = 16s^{\frac{4}{5}}$
* $10s^1 \cdot s^{-\frac{1}{5}} = 10s^{1 - \frac{1}{5}} = 10s^{\frac{5}{5} - \frac{1}{5}} = 10s^{\frac{4}{5}}$

Now substitute these back into the expression:
$3s^{\frac{4}{5}} + 16s^{\frac{4}{5}} - 5s^{\frac{11}{5}} + 10s^{\frac{4}{5}}$

Step 3: Combine like terms. The terms with exponent $\frac{4}{5}$ can be added together.
$(3 + 16 + 10)s^{\frac{4}{5}} - 5s^{\frac{11}{5}}$
$29s^{\frac{4}{5}} - 5s^{\frac{11}{5}}$

Step 4: Factor out the term with the smallest exponent, which is $s^{\frac{4}{5}}$.
$s^{\frac{4}{5}} (29 - 5s^{\frac{11}{5} - \frac{4}{5}})$
$s^{\frac{4}{5}} (29 - 5s^{\frac{7}{5}})$

Step 5: Convert back to radical notation.
$\sqrt[5]{s^4}(29 - 5\sqrt[5]{s^7})$

---

2) Simplify: $3g \cdot \sqrt[5]{20} \cdot m^{-8} \cdot g^{-3} \cdot \sqrt[5]{5} \cdot m^{-5}$

Step 1: Group the variables and numbers together.
$(3 \cdot g \cdot g^{-3}) \cdot (\sqrt[5]{20} \cdot \sqrt[5]{5}) \cdot (m^{-8} \cdot m^{-5})$

Step 2: Simplify the $g$ terms using exponent rules ($g^1 \cdot g^{-3} = g^{-2}$).
$3g^{-2}$

Step 3: Simplify the radicals. Since they have the same index (5), we can multiply the numbers inside.
$\sqrt[5]{20 \cdot 5} = \sqrt[5]{100}$

Step 4: Simplify the $m$ terms ($m^{-8} \cdot m^{-5} = m^{-13}$).
$m^{-13}$

Step 5: Combine everything.
$3g^{-2} \cdot \sqrt[5]{100} \cdot m^{-13}$

Step 6: Rewrite with positive exponents by moving negative exponents to the denominator.
$\frac{3\sqrt[5]{100}}{g^2 m^{13}}$

---

3) Simplify: $3m\sqrt[5]{63} \cdot m^{-4}\sqrt[5]{21} \cdot m^{-4}$

Step 1: Group the coefficients, variables, and radicals.
$(3 \cdot m \cdot m^{-4} \cdot m^{-4}) \cdot (\sqrt[5]{63} \cdot \sqrt[5]{21})$

Step 2: Simplify the $m$ terms. Add the exponents: $1 + (-4) + (-4) = -7$.
$3m^{-7}$

Step 3: Simplify the radicals. Multiply the insides: $63 \cdot 21$.
Note that $63 = 9 \cdot 7$ and $21 = 3 \cdot 7$.
So, $63 \cdot 21 = (9 \cdot 7) \cdot (3 \cdot 7) = 27 \cdot 49 = 27 \cdot 7^2$.
$\sqrt[5]{27 \cdot 7^2} = \sqrt[5]{3^3 \cdot 7^2} = \sqrt[5]{1323}$
There are no perfect fifth powers in 1323 (since $2^5=32$ and $3^5=243$), so the radical stays as is.

Step 4: Combine the parts.
$3m^{-7}\sqrt[5]{1323}$

Step 5: Write with positive exponents.
$\frac{3\sqrt[5]{1323}}{m^7}$

---

4) Simplify: $3 \cdot a^4 \cdot b^4 \cdot \sqrt[5]{81} \cdot a \cdot a^4 \cdot b^4 \cdot \sqrt[5]{273} \cdot b^4$

*(Note: Based on standard problems of this type, $\sqrt[5]{273}$ is likely a typo for $\sqrt[5]{243}$ because $243 = 3^5$, which simplifies perfectly. I will solve assuming it is 243. If it is strictly 273, the radical cannot be simplified further.)*

Assumption: The term is $\sqrt[5]{243}$.

Step 1: Group like terms.
Variables $a$: $a^4 \cdot a^1 \cdot a^4 = a^{4+1+4} = a^9$
Variables $b$: $b^4 \cdot b^4 \cdot b^4 = b^{4+4+4} = b^{12}$
Coefficients/Radicals: $3 \cdot \sqrt[5]{81} \cdot \sqrt[5]{243}$

Step 2: Simplify the radicals.
$\sqrt[5]{243} = \sqrt[5]{3^5} = 3$
So the coefficient part becomes: $3 \cdot \sqrt[5]{81} \cdot 3 = 9\sqrt[5]{81}$

Step 3: Combine everything.
$9 a^9 b^{12} \sqrt[5]{81}$

*(If the number was strictly 273, the answer would be $3 a^9 b^{12} \sqrt[5]{22059}$)*. Given the context of school worksheets, the intended answer is almost certainly based on perfect powers.

Final Answer for 4: $9 a^9 b^{12} \sqrt[5]{81}$

---

5) Simplify: $-2s \cdot t \cdot b^4 \cdot \sqrt[5]{27} \cdot s^{-5} \cdot -5 \cdot s \cdot b^4 \cdot \sqrt[5]{27} \cdot s^{-6}$

Step 1: Group coefficients, variables, and radicals.
Coefficients: $(-2) \cdot (-5) = 10$
Radicals: $\sqrt[5]{27} \cdot \sqrt[5]{27} = \sqrt[5]{27^2} = \sqrt[5]{729}$
Variable $s$: $s^1 \cdot s^{-5} \cdot s^1 \cdot s^{-6} = s^{1-5+1-6} = s^{-9}$
Variable $t$: $t^1$
Variable $b$: $b^4 \cdot b^4 = b^8$

Step 2: Assemble the expression.
$10 \cdot s^{-9} \cdot t \cdot b^8 \cdot \sqrt[5]{729}$

Step 3: Check if $\sqrt[5]{729}$ simplifies.
$729 = 3^6$.
$\sqrt[5]{3^6} = 3^{\frac{6}{5}} = 3^1 \cdot 3^{\frac{1}{5}} = 3\sqrt[5]{3}$.

Step 4: Update the coefficient.
$10 \cdot 3\sqrt[5]{3} = 30\sqrt[5]{3}$

Step 5: Write final expression with positive exponents.
$\frac{30 t b^8 \sqrt[5]{3}}{s^9}$

---

6) Simplify: $-5 \cdot s \cdot t \cdot b^4 \cdot \sqrt[5]{27} \cdot s^{-5} \cdot -8 \cdot s \cdot b^4 \cdot \sqrt[5]{27} \cdot s^{-6}$

This problem is very similar to #5, just with different coefficients.

Step 1: Group coefficients.
$(-5) \cdot (-8) = 40$

Step 2: Group radicals.
$\sqrt[5]{27} \cdot \sqrt[5]{27} = \sqrt[5]{729} = 3\sqrt[5]{3}$ (as calculated in problem 5).

Step 3: Group variables.
$s$: $s^1 \cdot s^{-5} \cdot s^1 \cdot s^{-6} = s^{-9}$
$t$: $t^1$
$b$: $b^4 \cdot b^4 = b^8$

Step 4: Combine coefficients.
$40 \cdot 3\sqrt[5]{3} = 120\sqrt[5]{3}$

Step 5: Write final expression with positive exponents.
$\frac{120 t b^8 \sqrt[5]{3}}{s^9}$

──────────────────────────────────────

Final Answer:

1) $\sqrt[5]{s^4}(29 - 5\sqrt[5]{s^7})$ or $29s^{\frac{4}{5}} - 5s^{\frac{11}{5}}$

2) $\frac{3\sqrt[5]{100}}{g^2 m^{13}}$

3) $\frac{3\sqrt[5]{1323}}{m^7}$

4) $9 a^9 b^{12} \sqrt[5]{81}$

5) $\frac{30 t b^8 \sqrt[5]{3}}{s^9}$

6) $\frac{120 t b^8 \sqrt[5]{3}}{s^9}$