$$27+125y^3 = 0$$

$$ \begin{matrix}y_1 = -\dfrac{ 3 }{ 5 } & y_2 = \dfrac{ 3 }{ 10 }+\dfrac{ 3 \sqrt{ 3}}{ 10 }i & y_3 = \dfrac{ 3 }{ 10 }-\dfrac{ 3 \sqrt{ 3}}{ 10 }i \end{matrix} $$

$ \color{blue}{ 125x^{3}+27 } $ is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factors of the leading coefficient ( 125 ) are 1 5 25 125 .The factors of the constant term (27) are 1 3 9 27 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 25 } , ~ \pm \frac{ 3 }{ 125 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 5 } , ~ \pm \frac{ 9 }{ 25 } , ~ \pm \frac{ 9 }{ 125 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 5 } , ~ \pm \frac{ 27 }{ 25 } , ~ \pm \frac{ 27 }{ 125 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(-\frac{ 3 }{ 5 }) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{ 5 x + 3 } $

$$ \frac{ 125x^{3}+27 }{ \color{blue}{ 5x + 3 } } = 25x^{2}-15x+9 $$

Polynomial $ 25x^{2}-15x+9 $ can be used to find the remaining roots.

$ \color{blue}{ 25x^{2}-15x+9 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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