Cphill is amazing (NEED MORE HELP)

9^3k * 27^3k-1 * 3^k  all divided by 81^4k  like a fraction

$\;\;\;\frac{9^{3k} * 27^{3k-1} * 3^k }{ 81^{4k} }\\ =\frac{3^{2*3k} * 3^{3(3k-1)} * 3^k }{ 3^{4*4k} }\\ =\frac{3^{6k} * 3^{9k-3} * 3^k }{ 3^{16k} }\\ =3^{6k+9k-3+k-16k}\\ =3^{-3}\\ =\frac{1}{3^3}\\ =\frac{1}{27}$

[9^(3k) * 27^(3k-1) * 3^k] /   81^(4k)

[ (3^2)^(3k) * (3^3)^(3k - 1) * 3^k] / [3^4]^{4k)

[ 3^(6k) * (3)^(9k - 3) * 3^k ]  / [3^16]^k

[  3^ ( 6k + 9k - 3 + k ) ] / [3^(16k)]

[ 3^(16k - 3) ] / [3^ (16k)]

3^(16k - 3 - 16k)  =

3^( -3 )

1 / 27

Rats  !!!.....Melody beat me to it.....!!!!!

Now Chris,

You would have seen me working on it before you even started your answer :/

Nah.....I think you "stealth" posted it.....LOL!!!!!!

Is this a math or a drama website?