Answer:
No.
Step-by-step explanation:
[tex]\sqrt[3]{s^{64}}[/tex]
[tex]\left(s^{64}\right)^{\frac{1}{3}}[/tex]
[tex]s^{\frac{64}{3}[/tex]
[tex]s^{21\frac{1}{3}}[/tex]
Answer:
no
Step-by-step explanation:
it is not a perfect cube
although 64 (4^3)is a perfect cube, s is not
