The numbers [tex]125a^6[/tex], [tex]-64q^{12}[/tex], and [tex]-c^{30}d^9f^{18}g^3[/tex] are perfact cubes as their cuberoot can be found, and the other number are not.
The given numbers are [tex]-3, \; 125a^6,\; 8m^{20}, \; 100w^{24}x^3,\; -64q^{12},\; -c^{30}d^9f^{18}g^3[/tex].
Now, the numbers which will have perfect cube root will be perfect cubes.
Consider the number [tex]-3[/tex]. It is not a cube of any number.
Consider the number [tex]125a^6[/tex]. Now, the cube root of the number will be,
[tex]125a^6=5^3a^6\\\sqrt[3]{125a^6} =\sqrt[3]{5^3a^6} \\\sqrt[3]{125a^6}=5\times a^2[/tex]
So, the number [tex]125a^6[/tex] is a perfect cube.
Consider the number [tex]8m^{20}[/tex]. Now, the cube root of the number will be,
[tex]8m^{20}=2^3m^{20}\\\sqrt[3]{8m^{20}} =\sqrt[3]{2^3m^{20}} \\\sqrt[3]{8m^{20}}=2\times m^{20/3}[/tex]
So, the number [tex]8m^{20}[/tex] is not a perfect cube.
Consider the number [tex]100w^{24}x^3[/tex]. Now, the cube root of the number will be,
[tex]100w^{24}x^3=10^2w^{24}x^3\\\sqrt[3]{100w^{24}x^3} =\sqrt[3]{10^2w^{24}x^3} \\\sqrt[3]{100w^{24}x^3}=10^{2/3}\times w^8x[/tex]
So, the number [tex]100w^{24}x^3[/tex] is not a perfect cube.
Consider the number [tex]-64q^{12}[/tex]. Now, the cube root of the number will be,
[tex]-64q^{12}=(-4)^3 q^{12}\\\sqrt[3]{-64q^{12}} =\sqrt[3]{(-4)^3 q^{12}} \\\sqrt[3]{-64q^{12}}=(-4)\times q^4[/tex]
So, the number [tex]-64q^{12}[/tex] is a perfect cube.
Consider the number [tex]-c^{30}d^9f^{18}g^3[/tex]. Now, the cube root of the number will be,
[tex]-c^{30}d^9f^{18}g^3=(-1)^3c^{30}d^9f^{18}g^3\\\sqrt[3]{-c^{30}d^9f^{18}g^3} =\sqrt[3]{(-1)^3c^{30}d^9f^{18}g^3} \\\sqrt[3]{-c^{30}d^9f^{18}g^3}=(-1)c^{10}d^3f^6g[/tex]
So, the number [tex]-c^{30}d^9f^{18}g^3[/tex] is a perfect cube.
Therefore, the numbers [tex]125a^6[/tex], [tex]-64q^{12}[/tex], and [tex]-c^{30}d^9f^{18}g^3[/tex] are perfact cubes and the other number are not.
For more details, refer the link:
https://brainly.com/question/16925380
