Please answer the two questions!

Given:
The expressions are
(c) [tex]\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3[/tex]
(d) [tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}[/tex]
To find:
The simplified form of the given expression.
Solution:
(c)
We have,
[tex]\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3[/tex]
We know that, zero to the power of a non-zero number is always 1. So, [tex]\left(\dfrac{2^4\times 3^6}{12^2}\right)^0=1[/tex]
[tex]\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=(1)^3[/tex]
[tex]\left\{\left(\dfrac{2^4\times 3^6}{12^2}\right)^0\right\}^3=1[/tex]
Therefore, the value of the given expression is 1.
(d)
We have,
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}[/tex]
It can be written as
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13^3\times 1}{(65\times 49)^2}[/tex]
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13\times 13\times 13}{(65\times 49)(65\times 49)}[/tex]
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{(5\times 49)(5\times 49)}[/tex]
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}[/tex]
[tex]\dfrac{13^3\times 7^0}{\{(65\times 49)^2\}^1}=\dfrac{13}{60025}[/tex]
Therefore, the value of given expression is [tex]\dfrac{13}{60025}[/tex].