How can I solve (27a^-3b^6c^3)^1/3?

.. (27 a^-3 b^6 c^3)^1/3
= (27)^1/3 (a^-3)^1/3 (b^6)^1/3 (c^3)^1/3
= 3 (a^-1) (b^2) (c^1)
= 3b²c/a

3b^2c/a

(27a^-3b^6c^3)^1/3
=27^(1/3) a^-3*(1/3) b^6*(1/3) c^3*(1/3) using (a^x)^y = a^(xy) and (ab)^x=a^x b^x
=3a^(-1) b^2 c
3 b^2 c
= -----------
a

(3/a)b^2c