Question

1. $15sqrt{2} cdot -2sqrt{20}$
2. $4sqrt3{-9} cdot 7sqrt3{48}$
3. $3sqrt{18w^7} cdot 10sqrt{4w^9}$
4. $sqrt4{8a^3b} cdot sqrt4{10a^2b^7}$
5. $2sqrt{3}(sqrt{27} - 8sqrt{6})$
6. $(5sqrt{5} - 2)(sqrt{5} - 1)$
7. $(10 + 3sqrt{7})(10 - 3sqrt{7})$
8. $(3 - 2sqrt{k})^2$

Explanation:

Let's solve each problem one by one:

Problem 5: \( 15\sqrt{2} \cdot -2\sqrt{20} \)

Step 1: Multiply the coefficients and the radicals separately.

First, multiply the coefficients: \( 15 \times (-2) = -30 \).
Then, multiply the radicals: \( \sqrt{2} \times \sqrt{20} = \sqrt{2 \times 20} = \sqrt{40} \).

Step 2: Simplify \( \sqrt{40} \).

\( \sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10} \).

Step 3: Multiply the results from Step 1 and Step 2.

\( -30 \times 2\sqrt{10} = -60\sqrt{10} \).

Step 1: Multiply the coefficients and the radicals separately.

Coefficients: \( 4 \times 7 = 28 \).
Radicals: \( \sqrt[3]{-9} \times \sqrt[3]{48} = \sqrt[3]{-9 \times 48} = \sqrt[3]{-432} \).

Step 2: Simplify \( \sqrt[3]{-432} \).

Factor \( -432 \): \( -432 = -8 \times 54 = -8 \times 27 \times 2 \).
So, \( \sqrt[3]{-432} = \sqrt[3]{-8 \times 27 \times 2} = \sqrt[3]{-8} \times \sqrt[3]{27} \times \sqrt[3]{2} = -2 \times 3 \times \sqrt[3]{2} = -6\sqrt[3]{2} \).

Step 3: Multiply the results from Step 1 and Step 2.

\( 28 \times (-6\sqrt[3]{2}) = -168\sqrt[3]{2} \).

Step 1: Multiply the coefficients and the radicals separately.

Coefficients: \( 3 \times 10 = 30 \).
Radicals: \( \sqrt{18w^7} \times \sqrt{4w^9} = \sqrt{18w^7 \times 4w^9} = \sqrt{72w^{16}} \).

Step 2: Simplify \( \sqrt{72w^{16}} \).

Factor \( 72 \): \( 72 = 36 \times 2 \).
\( w^{16} = (w^8)^2 \).
So, \( \sqrt{72w^{16}} = \sqrt{36 \times 2 \times (w^8)^2} = \sqrt{36} \times \sqrt{(w^8)^2} \times \sqrt{2} = 6w^8\sqrt{2} \).

Step 3: Multiply the results from Step 1 and Step 2.

\( 30 \times 6w^8\sqrt{2} = 180w^8\sqrt{2} \).

Answer:

\( -60\sqrt{10} \)

Problem 6: \( 4\sqrt[3]{-9} \cdot 7\sqrt[3]{48} \)