Question

1. $\left(4w^{2}\cdot2w\cdot w^{3}\

ight)^{2}$

1. $\left(7k^{6}z\

ight)^{2}$

1. $\left(3z^{3}\cdot4z\

ight)^{2}$

1. $\left(b\cdot3b^{3}\

ight)^{2}$

Explanation:

Problem 8:

Step1: Multiply coefficients and combine exponents of \( w \) inside the parentheses.

First, multiply the coefficients: \( 4 \cdot 2 = 8 \). Then, for the variable \( w \), use the rule \( a^m \cdot a^n = a^{m + n} \). So \( w^2 \cdot w \cdot w^3 = w^{2 + 1 + 3} = w^6 \). So inside the parentheses, we have \( 8w^6 \).

Step2: Apply the exponent outside the parentheses.

Now, raise \( 8w^6 \) to the power of 2. Using the rule \( (ab)^n = a^n b^n \), we get \( 8^2 \cdot (w^6)^2 \). Calculate \( 8^2 = 64 \) and \( (w^6)^2 = w^{6 \cdot 2} = w^{12} \) (using the rule \( (a^m)^n = a^{m \cdot n} \)).

Step1: Apply the exponent to each factor inside the parentheses.

Using the rule \( (ab)^n = a^n b^n \), we raise 7, \( k^6 \), and \( z \) to the power of 2.

Step2: Calculate each part.

For the coefficient: \( 7^2 = 49 \). For \( k^6 \): \( (k^6)^2 = k^{6 \cdot 2} = k^{12} \) (using \( (a^m)^n = a^{m \cdot n} \)). For \( z \): \( z^2 \) (since \( z = z^1 \), so \( (z^1)^2 = z^{2} \)). Then multiply them together: \( 49 \cdot k^{12} \cdot z^2 = 49k^{12}z^2 \).

Step1: Multiply coefficients and combine exponents of \( z \) inside the parentheses.

Multiply the coefficients: \( 3 \cdot 4 = 12 \). For the variable \( z \), use \( z^3 \cdot z = z^{3 + 1} = z^4 \) (by \( a^m \cdot a^n = a^{m + n} \)). So inside the parentheses, we have \( 12z^4 \).

Step2: Apply the exponent outside the parentheses.

Raise \( 12z^4 \) to the power of 2. Using \( (ab)^n = a^n b^n \), we get \( 12^2 \cdot (z^4)^2 \). Calculate \( 12^2 = 144 \) and \( (z^4)^2 = z^{4 \cdot 2} = z^8 \) (using \( (a^m)^n = a^{m \cdot n} \)).

Answer:

\( 64w^{12} \)

Problem 9: