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
