$$10460353203 = (3x)^3$$
$$ \begin{matrix}x_1 = 729 & x_2 = -\dfrac{ 729 }{ 2 }+\dfrac{ 729 \sqrt{ 3}}{ 2 }i & x_3 = -\dfrac{ 729 }{ 2 }-\dfrac{ 729 \sqrt{ 3}}{ 2 }i \end{matrix} $$
$ \color{blue}{ -27x^{3}+10460353203 } $ 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 ( -27 ) are 1 3 9 27 .The factors of the constant term (10460353203) are 1 3 9 27 81 243 729 2187 6561 19683 59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 27 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 9 } , ~ \pm \frac{ 3 }{ 27 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 3 } , ~ \pm \frac{ 9 }{ 9 } , ~ \pm \frac{ 9 }{ 27 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 3 } , ~ \pm \frac{ 27 }{ 9 } , ~ \pm \frac{ 27 }{ 27 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 81 }{ 3 } , ~ \pm \frac{ 81 }{ 9 } , ~ \pm \frac{ 81 }{ 27 } , ~ \pm \frac{ 243 }{ 1 } , ~ \pm \frac{ 243 }{ 3 } , ~ \pm \frac{ 243 }{ 9 } , ~ \pm \frac{ 243 }{ 27 } , ~ \pm \frac{ 729 }{ 1 } , ~ \pm \frac{ 729 }{ 3 } , ~ \pm \frac{ 729 }{ 9 } , ~ \pm \frac{ 729 }{ 27 } , ~ \pm \frac{ 2187 }{ 1 } , ~ \pm \frac{ 2187 }{ 3 } , ~ \pm \frac{ 2187 }{ 9 } , ~ \pm \frac{ 2187 }{ 27 } , ~ \pm \frac{ 6561 }{ 1 } , ~ \pm \frac{ 6561 }{ 3 } , ~ \pm \frac{ 6561 }{ 9 } , ~ \pm \frac{ 6561 }{ 27 } , ~ \pm \frac{ 19683 }{ 1 } , ~ \pm \frac{ 19683 }{ 3 } , ~ \pm \frac{ 19683 }{ 9 } , ~ \pm \frac{ 19683 }{ 27 } , ~ \pm \frac{ 59049 }{ 1 } , ~ \pm \frac{ 59049 }{ 3 } , ~ \pm \frac{ 59049 }{ 9 } , ~ \pm \frac{ 59049 }{ 27 } , ~ \pm \frac{ 177147 }{ 1 } , ~ \pm \frac{ 177147 }{ 3 } , ~ \pm \frac{ 177147 }{ 9 } , ~ \pm \frac{ 177147 }{ 27 } , ~ \pm \frac{ 531441 }{ 1 } , ~ \pm \frac{ 531441 }{ 3 } , ~ \pm \frac{ 531441 }{ 9 } , ~ \pm \frac{ 531441 }{ 27 } , ~ \pm \frac{ 1594323 }{ 1 } , ~ \pm \frac{ 1594323 }{ 3 } , ~ \pm \frac{ 1594323 }{ 9 } , ~ \pm \frac{ 1594323 }{ 27 } , ~ \pm \frac{ 4782969 }{ 1 } , ~ \pm \frac{ 4782969 }{ 3 } , ~ \pm \frac{ 4782969 }{ 9 } , ~ \pm \frac{ 4782969 }{ 27 } , ~ \pm \frac{ 14348907 }{ 1 } , ~ \pm \frac{ 14348907 }{ 3 } , ~ \pm \frac{ 14348907 }{ 9 } , ~ \pm \frac{ 14348907 }{ 27 } , ~ \pm \frac{ 43046721 }{ 1 } , ~ \pm \frac{ 43046721 }{ 3 } , ~ \pm \frac{ 43046721 }{ 9 } , ~ \pm \frac{ 43046721 }{ 27 } , ~ \pm \frac{ 129140163 }{ 1 } , ~ \pm \frac{ 129140163 }{ 3 } , ~ \pm \frac{ 129140163 }{ 9 } , ~ \pm \frac{ 129140163 }{ 27 } , ~ \pm \frac{ 387420489 }{ 1 } , ~ \pm \frac{ 387420489 }{ 3 } , ~ \pm \frac{ 387420489 }{ 9 } , ~ \pm \frac{ 387420489 }{ 27 } , ~ \pm \frac{ 1162261467 }{ 1 } , ~ \pm \frac{ 1162261467 }{ 3 } , ~ \pm \frac{ 1162261467 }{ 9 } , ~ \pm \frac{ 1162261467 }{ 27 } , ~ \pm \frac{ 3486784401 }{ 1 } , ~ \pm \frac{ 3486784401 }{ 3 } , ~ \pm \frac{ 3486784401 }{ 9 } , ~ \pm \frac{ 3486784401 }{ 27 } , ~ \pm \frac{ 10460353203 }{ 1 } , ~ \pm \frac{ 10460353203 }{ 3 } , ~ \pm \frac{ 10460353203 }{ 9 } , ~ \pm \frac{ 10460353203 }{ 27 } ~ $$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(729) = 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}{x - 729} $
$$ \frac{ -27x^{3}+10460353203 }{ \color{blue}{ x - 729 } } = -27x^{2}-19683x-14348907 $$Polynomial $ -27x^{2}-19683x-14348907 $ can be used to find the remaining roots.
$ \color{blue}{ -27x^{2}-19683x-14348907 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.