Answer:
No
Reasoning:
If something is a perfect cube, it is able to be put under a cube root ([tex]\sqrt[3]{..}[/tex]) and will result in an integer (a non-decimal number > 0, basically).
So let's calculate [tex]\sqrt[3]{48}[/tex], and see if the result is an integer.
[tex]\sqrt[3]{48}[/tex] = 3.634.......
As you can see, the result is not an integer, therefore 48 is not a perfect cube.
