simplifiy unsing indices

Question

((3a^2b^2c^2+2abc)(2a^-1b^-1c^-1))^3

Answer 1

((3a^2 b^2 c^2 + 2abc)(2a^-1 b^-1 c^-1))^3
= [(3a^2 b^2 c^2)(2a^-1 b^-1 c^-1) + (2abc)(2a^-1 b^-1 c^-1)]^3
= [((3 * 2)a^(2 - 1) b^(2 - 1) c^(2 - 1)) + ((2 * 2)a^(1 - 1) b^(1 - 1) c^(1 - 1))]^3
= [6a^1 b^1 c^1 + 4a^0 b^0 c^0]^3
= [6abc + 4(1)(1)(1)]^3
= [6abc + 4]^3

Not sure if you need this simplified any further.

Answer 2

simply 1/4 - 1=