Question

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

ight)^{2}$

1. $\left(8k^{3}\

ight)^{2}$

1. $\left(9rn^{3}\

ight)^{3}$

1. $\left(3n\cdot 2n^{3}\cdot n^{3}\

ight)^{2}$

Explanation:

Problem 2:

Step1: Multiply coefficients and combine like terms

First, multiply the coefficients \(2\) and \(4\), and combine the \(c\) terms using the rule \(a^m \cdot a^n = a^{m + n}\). So \(2c^2 \cdot c^3 \cdot 4c=(2\times4)\times(c^{2 + 3+1}) = 8c^6\)

Step2: Apply the power of a power rule

Now, apply \((a^m)^n=a^{m\times n}\) to \((8c^6)^2\). We get \(8^2\times(c^{6\times2})=64c^{12}\)

Step1: Apply the power of a product rule

Using \((ab)^n=a^n\times b^n\), for \((8k^3)^2\), we have \(8^2\times(k^3)^2\)

Step2: Simplify each part

Calculate \(8^2 = 64\) and \((k^3)^2=k^{3\times2}=k^6\). So the result is \(64k^6\)

Step1: Apply the power of a product rule

For \((9rn^3)^3\), using \((ab)^n=a^n\times b^n\), we get \(9^3\times r^3\times(n^3)^3\)

Step2: Simplify each part

Calculate \(9^3=729\), \((n^3)^3 = n^{3\times3}=n^9\). So the result is \(729r^3n^9\)

Answer:

\(64c^{12}\)

Problem 3: