Properties of Exponents

Simplify the following expressions.

\begin{tabular}{|l|l|}
\hline
Original Expression & Simplified Expression \\
\hline
[tex]$6x\left(2x^2\right)^3$[/tex] & [tex]$\square$[/tex] \\
\hline
[tex]$\left(-4z^3\right)^2$[/tex] & [tex]$\square$[/tex] \\
\hline
[tex]$\left(2n^6\right)^3 \cdot \left(4n^7\right)^2$[/tex] & [tex]$\square$[/tex] \\
\hline
[tex]$2y \cdot 3y \cdot (7y)^2$[/tex] & [tex]$\square$[/tex] \\
\hline
\end{tabular}



Answer :

Sure! Let's simplify each expression step-by-step using the properties of exponents:

### Expression 1: [tex]\(6x(2x^2)^3\)[/tex]

1. First, simplify the exponent inside the parentheses: [tex]\((2x^2)^3\)[/tex].
- [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex].
- So, [tex]\((2x^2)^3 = 2^3 \cdot (x^2)^3\)[/tex].

2. Simplify further:
- [tex]\(2^3 = 8\)[/tex].
- [tex]\((x^2)^3 = x^{2 \cdot 3} = x^6\)[/tex].
- Hence, [tex]\((2x^2)^3 = 8x^6\)[/tex].

3. Now, multiply [tex]\(6x\)[/tex] by the simplified result:
- [tex]\(6x \cdot 8x^6 = 6 \cdot 8 \cdot x \cdot x^6\)[/tex].
- [tex]\(6 \cdot 8 = 48\)[/tex].
- [tex]\(x \cdot x^6 = x^{1 + 6} = x^7\)[/tex].

4. Thus, [tex]\(6x(2x^2)^3\)[/tex] simplifies to:
- [tex]\(\boxed{48x^7}\)[/tex].

### Expression 2: [tex]\((-4z^3)^2\)[/tex]

1. Simplify the exponent:
- Using [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex]:
- [tex]\((-4z^3)^2 = (-4)^2 \cdot (z^3)^2)\)[/tex].

2. Further simplification:
- [tex]\((-4)^2 = 16\)[/tex].
- [tex]\((z^3)^2 = z^{3 \cdot 2} = z^6\)[/tex].

3. Thus, [tex]\((-4z^3)^2\)[/tex] simplifies to:
- [tex]\(\boxed{16z^6}\)[/tex].

### Expression 3: [tex]\((2n^6)^3 \cdot (4n^7)^2\)[/tex]

1. Simplify each part separately:
- [tex]\((2n^6)^3\)[/tex]:
- Using [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex]:
- [tex]\((2^3) \cdot (n^6)^3 = 2^3 \cdot n^{6 \cdot 3}\)[/tex].
- [tex]\(2^3 = 8\)[/tex].
- [tex]\(n^{6 \cdot 3} = n^{18}\)[/tex].
- So, [tex]\((2n^6)^3 = 8n^{18}\)[/tex].

- [tex]\((4n^7)^2\)[/tex]:
- Using [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex]:
- [tex]\((4^2) \cdot (n^7)^2 = 4^2 \cdot n^{7 \cdot 2}\)[/tex].
- [tex]\(4^2 = 16\)[/tex].
- [tex]\(n^{7 \cdot 2} = n^{14}\)[/tex].
- So, [tex]\((4n^7)^2 = 16n^{14}\)[/tex].

2. Multiply the simplified parts together:
- [tex]\(8n^{18} \cdot 16n^{14}\)[/tex].
- [tex]\(8 \cdot 16 = 128\)[/tex].
- [tex]\(n^{18} \cdot n^{14} = n^{18 + 14} = n^{32}\)[/tex].

3. Thus, [tex]\((2n^6)^3 \cdot (4n^7)^2\)[/tex] simplifies to:
- [tex]\(\boxed{128n^{32}}\)[/tex].

### Expression 4: [tex]\(2y \cdot 3y \cdot (7y)^2\)[/tex]

1. First, simplify [tex]\((7y)^2\)[/tex]:
- Using [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex]:
- [tex]\((7y)^2 = 7^2 \cdot y^2\)[/tex].
- [tex]\(7^2 = 49\)[/tex].
- So, [tex]\((7y)^2 = 49y^2\)[/tex].

2. Multiply all parts together:
- [tex]\(2y \cdot 3y \cdot 49y^2\)[/tex].
- Combine the scalar multipliers: [tex]\(2 \cdot 3 \cdot 49 = 294\)[/tex].
- Combine the [tex]\(y\)[/tex] terms: [tex]\(y \cdot y \cdot y^2 = y^{1+1+2} = y^4\)[/tex].

3. Thus, [tex]\(2y \cdot 3y \cdot (7y)^2\)[/tex] simplifies to:
- [tex]\(\boxed{294y^4}\)[/tex].

The complete table with the simplified expressions would look like:

[tex]\[ \begin{tabular}{|l|l|} \hline \hline Original Expression & Simplified Expression \\ \hline \hline 6x(2x^2)^3 & 48x^7 \\ \hline (-4z^3)^2 & 16z^6 \\ \hline (2n^6)^3 \cdot (4n^7)^2 & 128n^{32} \\ \hline 2y \cdot 3y \cdot (7y)^2 & 294y^4 \\ \hline \end{tabular} \][/tex]