can i have the x,y,z for this question? THanks!

Answer:
x=4, y=-2, p=14
Step-by-step explanation:
First simplify the LHS:
[tex]LHS = (16p^8)^\frac{3}{2} \times (216p^{-3})^{-\frac{2}{3}}[/tex]
[tex]= 64 p^{12} \times \frac{1}{36} p^{2}[/tex]
[tex]=\frac{16}{9}p^{14}[/tex]
[tex]=\frac{2^4}{3^2}p^14\\=2^43^{-2}p^{14}[/tex]
Hence x=4, y=2, p=14