Algebra question

Find all real numbers x such that

$(3^x - 27)^3 + (27x - 3)^3 = (3x + 27x - 30)^3$

well, this is going to be a pain to expand, but i see no other way... Oh god.

Okay after 30 minutes of backbending expanding, i got this:

$27^x-9^{2+x}+3^{x+7}-19683 + 19683x^3-6561x^2+729x-27 = 27000x^3-81000x^2+81000x-27000$

yeah.... thats, quite a bit of math. Now lets simplify! YAY! *sigh..*

$27^x-9^{2+x}+3^{x+7} - 7317x^3 + 74439x^2 + 80271x = 0$

Im acually having fun doing this! No sarcasm like acually!

Okay lets acually do it another way.

$[ (3x - 27 + 27x - 3) ] [ (3x - 27)^2 - (3x - 27)(27x - 3) + ( 27x - 3)^2 ] = 27000x^3 - 81000x^2 + 81000x - 27000$

$( 30x - 30) [ (3(x - 9))^2 - ( 3 (x - 9) * 3(9x - 1) ) + (3 (9x - 1))^2 ] = the thing idon't wanna write$

$(30x - 30) [ 9 ( x^2 - 18x + 81) - ( 9(9x^2 - 82x + 9)) + (9 ( 81x^2 - 18x + 1)) = againidon't wannawriteit$

$9 (30x - 30) [ x^2 - 18x + 81 - 9x^2 + 82x - 9 + 81x^2 - 18x + 1 ] = yeahyeahyeah$

$270 ( x - 1) [ 73x^2 + 46x + 73 ] = ohgod$

the discriminant of $[ 73x^2 + 46x + 73 ]$is negative, so it has no real roots, just saying, you know.

so basically $270 ( x - 1) = (30x-30)^3$

more simplifying....

$270 ( x - 1) = 270(100x^3 - 300x^2 + 300x - 100)$

$x-1=100x^3 - 300x^2+300x-100$

$100x^3 - 300x^2 + 299x - 99 = 0$

i think ill let you take it from here

Note: I've spent more than an hour on this prob, but i haven't solved it yet, anybody that can give the solution i will appreciate it

can anybody online check my work? My brain is damaged.

Where do i go? to the left, where nothing is right, or to the right, where nothing is left?

Ahh i see... thank you for clarifying, i understand perfectly now