The maximum value of `8*27^(log_6x) + 27*8^(log_6 x) - x^3` has to be determined.

Let `y = log_6x^3`

`y = log_6x^3`

`=gt 6^y = x^3`

`8*27^(log_6x) + 27*8^(log_6 x) - x^3`

`= 8*(3^3)^(log_6x)+27*(2^3)^(log_6x) - x^3`

`= 8*3^(3*log_6x)+27*2^(3*log_6x) - x^3`

`=8*3^(log_6x^3)+27*2^(log_6x^3) - x^3`

`=8*3^y+27*2^y-6^y`

`=8*3^y+27*2^y-2^y*3^y-2^y*3^y-8*27+8*27`

`=8(3^y-27)-2^y(3^y-27)+8*27`

`=(8-2^y)(3^y-27)+8*27`.

The maximum value of `(8-2^y)(3^y-27)` is 0.

Therefore the maximum value of `(8-2^y)(3^y-27)+8*27` is `8*27 ` or 216.