Question

1. $(isqrt{3})^2 cdot (-8i)^2$
2. $sqrt{-24} cdot sqrt{-3} cdot sqrt{-2}$

Explanation:

Problem 20: \((i\sqrt{3})^2 \cdot (-8i)^2\)

Step 1: Simplify \((i\sqrt{3})^2\)

Using the power of a product rule \((ab)^n = a^n b^n\), we have \((i\sqrt{3})^2 = i^2 \cdot (\sqrt{3})^2\). We know that \(i^2 = -1\) and \((\sqrt{3})^2 = 3\), so substituting these values in, we get \(i^2 \cdot (\sqrt{3})^2 = -1 \cdot 3 = -3\).

Step 2: Simplify \((-8i)^2\)

Again, using the power of a product rule \((ab)^n = a^n b^n\), we have \((-8i)^2 = (-8)^2 \cdot i^2\). Calculating \((-8)^2 = 64\) and using \(i^2 = -1\), we get \((-8)^2 \cdot i^2 = 64 \cdot (-1) = -64\).

Step 3: Multiply the two results

Now we multiply the results from Step 1 and Step 2: \(-3 \cdot (-64) = 192\).

Step 1: Rewrite square roots of negative numbers

Recall that for any positive real number \(a\), \(\sqrt{-a} = i\sqrt{a}\). So we can rewrite each square root:

• \(\sqrt{-24} = i\sqrt{24}\)
• \(\sqrt{-3} = i\sqrt{3}\)
• \(\sqrt{-2} = i\sqrt{2}\)

Step 2: Multiply the imaginary units and the square roots

First, multiply the imaginary units: \(i \cdot i \cdot i = i^3\). We know that \(i^3 = i^2 \cdot i = -1 \cdot i = -i\).
Next, multiply the square roots: \(\sqrt{24} \cdot \sqrt{3} \cdot \sqrt{2}\). Using the property \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\), we can combine them: \(\sqrt{24 \cdot 3 \cdot 2} = \sqrt{144}\). And \(\sqrt{144} = 12\).

Step 3: Combine the results

Now we multiply the result from the imaginary units and the result from the square roots: \(-i \cdot 12 = -12i\).

Answer:

\(192\)

Problem 22: \(\sqrt{-24} \cdot \sqrt{-3} \cdot \sqrt{-2}\)