I need to slove this using cubic formula and i don't know how.... (x^3+27)/(x+3)?

(x^3 + 27)/(x + 3)
= (x + 3)(x^2 - 3x + 9)/(x + 3) (<= factorization)
= (x^2 - 3x + 9)/1 (<= cancelled out x + 3)
= x^2 - 3x + 9

= (x³ + 27)/(x + 3)
= x² - 3x + 9

Answer: x² - 3x + 9

Using this formula to factor the numerator.
a3 + b3 = (a + b)(a2 – ab + b2)

You have.
(x³ +3³)/(x +3)

Factoring.
(x +3)(x² -3x +3²)/(x +3)

Eliminate element (x +3) by canceling out.

x² -3x +9 = 0, x² -3x = -9, x² -3x +9/4 = -9 +9/4,
(x -3/2)² = -27/4, √((x -3/2)²) = √(-9/4).

Even root principle. ±

x -3/2 = ±√9/2, x = (3±-9)/(2)

No "real" solution. Only if your dealing with complex "imaginary" numbers is there a answer, like "i".

a^3+b^3=(a+b)(a^2-ab+b^2)
a=x
b=3
x^3+27 = (x+3)(x^2-3x+9) / (x+3)
=x^2-3x+9
The (x+3) cancels out.