Solution:
Given:
[tex]\begin{gathered} \sqrt[]{-128} \\ \\ \text{also,} \\ i^2=-1 \end{gathered}[/tex]
To simplify, we break the radical down into bits.
[tex]\begin{gathered} \sqrt[]{-128}=\sqrt[]{-1}\times\sqrt[]{128} \\ To\text{ simplify }\sqrt[]{128}\text{ , we split it into a number that is a p}\operatorname{erf}ect\text{ square and another number,} \\ \text{Hence,} \\ \sqrt[]{128}=\sqrt[]{64\times2}=\sqrt[]{64}\times\sqrt[]{2} \\ \text{This means,} \\ \sqrt[]{-128}=\sqrt[]{-1}\times\sqrt[]{64}\times\sqrt[]{2} \\ \text{But recall that;} \\ i^2=-1 \\ \text{Taking the square root of both sides,} \\ i=\sqrt[]{-1} \\ \text{Substituting it in the expression below,} \\ \sqrt[]{-128}=\sqrt[]{-1}\times\sqrt[]{64}\times\sqrt[]{2} \\ \sqrt[]{-128}=i\times8\times\sqrt[]{2} \\ \sqrt[]{-128}=8i\sqrt[]{2} \end{gathered}[/tex]
Therefore, the simplified form is;
[tex]8i\sqrt[]{2}[/tex]
