tentukan nilai p yang memenuhi !

^3√1/729 = p²
1/9 = p²
p² = 1/9
p = √1/9
p = 1/3

Mapel : Matematika
Materi : Akar Pangkat

[tex] \sqrt[3]{ \frac{1}{729} }=p^2 \\ \\ p^2= \sqrt[3]{ \frac{1}{9^3} } \\ \\ p^2= \frac{1^{ \frac{1}{3}} }{9^{(3\times \frac{1}{3}) }} \\ \\ p^2= \frac{1}{9} \\ \\ p= \sqrt{ \frac{1}{9} } \\ \\ p=\frac{ \sqrt{1} }{ \sqrt{9} } \\ \\ p= \frac{1}{3} [/tex]

***Semoga Terbantu***