Find all real numbers x such that \(\left(\dfrac{x}{3}\right)^3-3x^2+81x-729=25+2(5)(3)+9.\)
(x/3)^3 -3x^2 + 81x - 729 = 25 + 2(5)(3) + 9
Simplify the right side and note that the left side can be factored as :
(x/3 - 9)^3 = 64 take the cube root of both sides
(x/3 - 9) = 4 add 9 to both sides
x/3 = 13 multiply both sides by 3
x = 39 .....this is the only real solution......