We are tasked with simplifying $4 \times \sqrt[3]{-2048a^3b^2c^8}$.

Step-by-step process:

1. Factor the constant term (-2048): $-2048 = -2^{11}$ We express the constant as powers of 2, because we will take cube roots.

2. Simplify the cube root: We now break down the expression under the cube root: [ \sqrt[3]{-2048a^3b^2c^8} = \sqrt[3]{-2^{11}a^3b^2c^8} ] We can separate the terms: [ = \sqrt[3]{-2^{11}} \times \sqrt[3]{a^3} \times \sqrt[3]{b^2} \times \sqrt[3]{c^8} ]

3. Simplify each component:

   • $\sqrt[3]{-2^{11}} = -2^3 \times \sqrt[3]{2^2} = -8 \times \sqrt[3]{4}$
   • $\sqrt[3]{a^3} = a$
   • $\sqrt[3]{b^2}$ cannot be simplified further.
   • $\sqrt[3]{c^8} = c^2 \times \sqrt[3]{c}$

   So, the expression becomes: [ -8 \times \sqrt[3]{4} \times a \times \sqrt[3]{b^2} \times c^2 \times \sqrt[3]{c} ]

4. Multiply by 4 (as given in the problem): Now, we multiply the whole expression by 4: [ 4 \times \left( -8 \times \sqrt[3]{4} \times a \times \sqrt[3]{b^2} \times c^2 \times \sqrt[3]{c} \right) ] $= -32a c^2 \times \sqrt[3]{4b^2c}$

Final simplified expression:

$-32ac^2 \times \sqrt[3]{4b^2c}$

Let me know if you want more details or have any questions!

Here are 5 related questions:

1. How do you simplify cube roots of higher powers?
2. How do you simplify roots involving negative numbers?
3. What is the cube root of a product of variables?
4. How do you deal with cube roots of constants like 8, 27, or 64?
5. How do fractional exponents relate to cube roots?

Tip: When simplifying roots, try to factor out perfect powers to make the simplification process easier.