Given,
The expression is,
[tex]\begin{gathered} x^3=216 \\ x^3-216=0 \end{gathered}[/tex]
The rational theorem tells that if the polynomial has the rational 0 then it must be a fraction p/q, where p is a factor of constant term and q is the factor of leading coefficient.
The constant term 216 with factors:
[tex]1,2,3,4,6,8,9,12,18,24,27,36,54,72,108\text{ }and\text{ }216.[/tex]
The leading coefficient is 1, with a single factor of 1.
[tex]\begin{gathered} \frac{p}{q}=\frac{factor\text{ of 216}}{factor\text{ of 1}}=\pm\frac{1}{1},\operatorname{\pm}\frac{2}{1},\operatorname{\pm}\frac{3}{1},\operatorname{\pm}\frac{4}{1},\operatorname{\pm}\frac{6}{1},\operatorname{\pm}\frac{8}{1},\operatorname{\pm}\frac{9}{1},\operatorname{\pm}\frac{12}{1}, \\ \operatorname{\pm}\frac{18}{1},\operatorname{\pm}\frac{24}{1},\operatorname{\pm}\frac{27}{1},\operatorname{\pm}\frac{36}{1},\operatorname{\pm}\frac{54}{1},\operatorname{\pm}\frac{72}{1},\operatorname{\pm}\frac{108}{1},\operatorname{\pm}\frac{216}{1} \end{gathered}[/tex]
Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
p(6)=0 so x=6 is a root of a polynomial p(x).
Using factor theorem to find the remaining roots,
[tex]\begin{gathered} \frac{x^3-216}{x-6}=x^2+6x+36 \\ By\text{ formula method,} \\ x=\frac{-6\pm\sqrt{36-144}}{2} \\ x=\frac{-6\pm\sqrt{108}}{2} \\ x=\frac{-6\pm6\sqrt{-3}}{2} \\ x=-3\pm3\sqrt{3i} \end{gathered}[/tex]
Hence, the roots are 6, -3-3sqrt(3)i, and -3+3sqrt(3)i. So, option A is correct.
