Question

1. $(-4i)(2i)(-9i)$
2. $(isqrt{3})^2 cdot (-8i)^2$
3. $sqrt{-24} cdot sqrt{-3} cdot sqrt{-2}$

Explanation:

Problem 18: $(-4i)(2i)(-9i)$

Step 1: Multiply the coefficients and the imaginary units

First, multiply the coefficients: $(-4) \times 2 \times (-9) = 72$. Then, multiply the imaginary units: $i \times i \times i = i^3$.
So, we have $72 \times i^3$.

Step 2: Simplify $i^3$

Recall that $i^2 = -1$, so $i^3 = i^2 \times i = -1 \times i = -i$.
Substitute back: $72 \times (-i) = -72i$.

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

Using the power of a product rule: $(ab)^n = a^n b^n$. So, $(i\sqrt{3})^2 = i^2 \times (\sqrt{3})^2$.
We know $i^2 = -1$ and $(\sqrt{3})^2 = 3$. So, $i^2 \times (\sqrt{3})^2 = -1 \times 3 = -3$.

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

Using the power of a product rule: $(-8i)^2 = (-8)^2 \times i^2$.
$(-8)^2 = 64$ and $i^2 = -1$. So, $(-8)^2 \times i^2 = 64 \times (-1) = -64$.

Step 3: Multiply the two results

Multiply $-3$ and $-64$: $(-3) \times (-64) = 192$.

Step 1: Rewrite square roots of negative numbers

Recall that $\sqrt{-a} = i\sqrt{a}$ for $a > 0$. So:
$\sqrt{-24} = i\sqrt{24}$, $\sqrt{-3} = i\sqrt{3}$, $\sqrt{-2} = i\sqrt{2}$.

Step 2: Multiply the imaginary units and the square roots

Multiply the imaginary units: $i \times i \times i = i^3$.
Multiply the square roots: $\sqrt{24} \times \sqrt{3} \times \sqrt{2}$.
First, simplify the product of square roots: $\sqrt{24 \times 3 \times 2} = \sqrt{144} = 12$.

Step 3: Simplify $i^3$ and multiply with 12

We know $i^3 = -i$ (from $i^3 = i^2 \times i = -1 \times i = -i$).
So, $i^3 \times 12 = -i \times 12 = -12i$.

Answer:

$-72i$

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