We can factor the left-hand side of the equation as follows:
(3x - 27)^3 + (27x - 3)^3 = ((3x - 27) + (27x - 3))(((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2)
= (30x - 30)((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2).
We can factor the right-hand side of the equation as follows:
(3x + 27x - 30)^3 = (60x - 30)^3.
Since (60x−30)3=(30x−30)((60x−30)2), we can write the equation as follows:
(30x - 30)((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2) = (60x - 30)((60x - 30)^2).
Since (30x−30)=0, we can cancel it from both sides of the equation, which gives us:
((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2) = (60x - 30)^2.
Expanding both sides of the equation, we get:
9x^2 - 162x + 729 - 27x^3 + 196x^2 - 171x + 243 + 729x^2 - 189x + 9 = 3600x^2 - 7200x + 2700
Combining like terms, we get:
2547x^2 - 2316x - 2507 = 0
This quadratic factors as follows:
(x - 1)(2547x + 2507) = 0
The only real solution to this equation is x=1