Question

1. \sqrt{-6} \cdot \sqrt{-12} \cdot \sqrt{-5}

Explanation:

Step1: Express using imaginary unit

Recall that \(\sqrt{-a} = i\sqrt{a}\) for \(a>0\), where \(i\) is the imaginary unit with \(i^2=-1\). So we rewrite each square root:
\(\sqrt{-6}=i\sqrt{6}\), \(\sqrt{-12}=i\sqrt{12}\), \(\sqrt{-5}=i\sqrt{5}\)

Step2: Multiply the imaginary units and the radicals

Multiply the three expressions together:
\((i\sqrt{6})\cdot(i\sqrt{12})\cdot(i\sqrt{5})\)
First, multiply the imaginary units: \(i\cdot i\cdot i = i^{3}\). Since \(i^{2}=-1\), then \(i^{3}=i^{2}\cdot i=-i\)
Next, multiply the radicals: \(\sqrt{6}\cdot\sqrt{12}\cdot\sqrt{5}\). We know that \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), so \(\sqrt{6}\cdot\sqrt{12}=\sqrt{6\times12}=\sqrt{72}\), and then \(\sqrt{72}\cdot\sqrt{5}=\sqrt{72\times5}=\sqrt{360}\)
Simplify \(\sqrt{360}\): \(\sqrt{36\times10}=6\sqrt{10}\)

Step3: Combine the results

Now combine the result from the imaginary units and the radicals:
\(-i\cdot6\sqrt{10}=-6i\sqrt{10}\)

Answer:

\(-6i\sqrt{10}\)