Answer:
[tex]\large\boxed{9x^8,\ 25x^{12},\ 36x^{16}}[/tex]
Step-by-step explanation:
[tex]Use\\(\sqrt{s})^2=s\ for\ s\geq0\\(a^n)^m=a^{nm}\\(ab)^n=a^nb^n\\a^n\cdot a^m=a^{n+m}\\===========================\\6x^2=(\sqrt6)^2x^2=(x\sqrt6)^2\to\boxed{NOT}\\\\9x^8=3^2x^{4\cdot2}=3^2(x^4)^2=(3x^4)^2\to\boxed{YES}\\\\16x^9=4^2x^{8+1}=4^2x^{4\cdot2}\cdot x=4^2(x^4)^2x=(4x^4)^2x\to\boxed{NOT}\\\\25x^{12}=5^2x^{6\cdot2}=5^2(x^6)^2=(5x^6)^2\to\boxed{YES}\\\\36x^{16}=6^2x^{8\cdot2}=6^2(x^8)^2=(6x^8)^2\to\boxed{YES}[/tex]
