Let's solve the expression $-3 \cdot \sqrt[3]{16} + \sqrt[3]{128}$.

Step 1: Simplify $\sqrt[3]{16}$

We know: $16 = 2^4$ So, [ \sqrt[3]{16} = \sqrt[3]{2^4} = 2^{\frac{4}{3}} ]

Step 2: Simplify $\sqrt[3]{128}$

We know: $128 = 2^7$ So, [ \sqrt[3]{128} = \sqrt[3]{2^7} = 2^{\frac{7}{3}} ]

Step 3: Substitute into the original expression

Now, substitute these into the original expression: $-3 \cdot 2^{\frac{4}{3}} + 2^{\frac{7}{3}}$

Step 4: Factor out $2^{\frac{4}{3}}$

We can factor out $2^{\frac{4}{3}}$ from both terms: $2^{\frac{4}{3}} \left( -3 + 2^{\frac{3}{3}} \right) = 2^{\frac{4}{3}} \left( -3 + 2 \right)$ This simplifies to: $2^{\frac{4}{3}} \cdot (-1) = -2^{\frac{4}{3}}$

Step 5: Match to the answer choices

We have simplified the expression to $-2^{\frac{4}{3}}$, which corresponds to the first option: $\boxed{-2^{\frac{4}{3}}}$

Therefore, the correct answer is the first option.

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How do you simplify cube roots of larger numbers?
2. Can cube roots be simplified similarly to square roots?
3. What happens when negative numbers are involved with cube roots?
4. How do you handle fractional exponents in more complex problems?
5. What is the general method for factoring expressions like this?

Tip: Remember, cube roots of powers of 2 can often be simplified by expressing the number as a power of 2 and applying fractional exponents.