$$1000x^3-729 = 0$$

$$ \begin{matrix}x_1 = \dfrac{ 9 }{ 10 } & x_2 = -\dfrac{ 9 }{ 20 }+\dfrac{ 9 \sqrt{ 3}}{ 20 }i & x_3 = -\dfrac{ 9 }{ 20 }-\dfrac{ 9 \sqrt{ 3}}{ 20 }i \end{matrix} $$

$ \color{blue}{ 1000x^{3}-729 } $ 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 ( 1000 ) are 1 2 4 5 8 10 20 25 40 50 100 125 200 250 500 1000 .The factors of the constant term (-729) are 1 3 9 27 81 243 729 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 1 }{ 25 } , ~ \pm \frac{ 1 }{ 40 } , ~ \pm \frac{ 1 }{ 50 } , ~ \pm \frac{ 1 }{ 100 } , ~ \pm \frac{ 1 }{ 125 } , ~ \pm \frac{ 1 }{ 200 } , ~ \pm \frac{ 1 }{ 250 } , ~ \pm \frac{ 1 }{ 500 } , ~ \pm \frac{ 1 }{ 1000 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 3 }{ 10 } , ~ \pm \frac{ 3 }{ 20 } , ~ \pm \frac{ 3 }{ 25 } , ~ \pm \frac{ 3 }{ 40 } , ~ \pm \frac{ 3 }{ 50 } , ~ \pm \frac{ 3 }{ 100 } , ~ \pm \frac{ 3 }{ 125 } , ~ \pm \frac{ 3 }{ 200 } , ~ \pm \frac{ 3 }{ 250 } , ~ \pm \frac{ 3 }{ 500 } , ~ \pm \frac{ 3 }{ 1000 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 9 }{ 5 } , ~ \pm \frac{ 9 }{ 8 } , ~ \pm \frac{ 9 }{ 10 } , ~ \pm \frac{ 9 }{ 20 } , ~ \pm \frac{ 9 }{ 25 } , ~ \pm \frac{ 9 }{ 40 } , ~ \pm \frac{ 9 }{ 50 } , ~ \pm \frac{ 9 }{ 100 } , ~ \pm \frac{ 9 }{ 125 } , ~ \pm \frac{ 9 }{ 200 } , ~ \pm \frac{ 9 }{ 250 } , ~ \pm \frac{ 9 }{ 500 } , ~ \pm \frac{ 9 }{ 1000 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 27 }{ 4 } , ~ \pm \frac{ 27 }{ 5 } , ~ \pm \frac{ 27 }{ 8 } , ~ \pm \frac{ 27 }{ 10 } , ~ \pm \frac{ 27 }{ 20 } , ~ \pm \frac{ 27 }{ 25 } , ~ \pm \frac{ 27 }{ 40 } , ~ \pm \frac{ 27 }{ 50 } , ~ \pm \frac{ 27 }{ 100 } , ~ \pm \frac{ 27 }{ 125 } , ~ \pm \frac{ 27 }{ 200 } , ~ \pm \frac{ 27 }{ 250 } , ~ \pm \frac{ 27 }{ 500 } , ~ \pm \frac{ 27 }{ 1000 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 81 }{ 2 } , ~ \pm \frac{ 81 }{ 4 } , ~ \pm \frac{ 81 }{ 5 } , ~ \pm \frac{ 81 }{ 8 } , ~ \pm \frac{ 81 }{ 10 } , ~ \pm \frac{ 81 }{ 20 } , ~ \pm \frac{ 81 }{ 25 } , ~ \pm \frac{ 81 }{ 40 } , ~ \pm \frac{ 81 }{ 50 } , ~ \pm \frac{ 81 }{ 100 } , ~ \pm \frac{ 81 }{ 125 } , ~ \pm \frac{ 81 }{ 200 } , ~ \pm \frac{ 81 }{ 250 } , ~ \pm \frac{ 81 }{ 500 } , ~ \pm \frac{ 81 }{ 1000 } , ~ \pm \frac{ 243 }{ 1 } , ~ \pm \frac{ 243 }{ 2 } , ~ \pm \frac{ 243 }{ 4 } , ~ \pm \frac{ 243 }{ 5 } , ~ \pm \frac{ 243 }{ 8 } , ~ \pm \frac{ 243 }{ 10 } , ~ \pm \frac{ 243 }{ 20 } , ~ \pm \frac{ 243 }{ 25 } , ~ \pm \frac{ 243 }{ 40 } , ~ \pm \frac{ 243 }{ 50 } , ~ \pm \frac{ 243 }{ 100 } , ~ \pm \frac{ 243 }{ 125 } , ~ \pm \frac{ 243 }{ 200 } , ~ \pm \frac{ 243 }{ 250 } , ~ \pm \frac{ 243 }{ 500 } , ~ \pm \frac{ 243 }{ 1000 } , ~ \pm \frac{ 729 }{ 1 } , ~ \pm \frac{ 729 }{ 2 } , ~ \pm \frac{ 729 }{ 4 } , ~ \pm \frac{ 729 }{ 5 } , ~ \pm \frac{ 729 }{ 8 } , ~ \pm \frac{ 729 }{ 10 } , ~ \pm \frac{ 729 }{ 20 } , ~ \pm \frac{ 729 }{ 25 } , ~ \pm \frac{ 729 }{ 40 } , ~ \pm \frac{ 729 }{ 50 } , ~ \pm \frac{ 729 }{ 100 } , ~ \pm \frac{ 729 }{ 125 } , ~ \pm \frac{ 729 }{ 200 } , ~ \pm \frac{ 729 }{ 250 } , ~ \pm \frac{ 729 }{ 500 } , ~ \pm \frac{ 729 }{ 1000 } ~ $$

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{ 9 }{ 10 }) = 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}{ 10 x - 9 } $

$$ \frac{ 1000x^{3}-729 }{ \color{blue}{ 10x - 9 } } = 100x^{2}+90x+81 $$

Polynomial $ 100x^{2}+90x+81 $ can be used to find the remaining roots.

$ \color{blue}{ 100x^{2}+90x+81 } $ 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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