3 x^{3}-7 x^{2}+27 x-63=0

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Step:1 ：We are given the cubic equation (3 x^{3}-7 x^{2}+27 x-63=0).Step:2 ：To solve it, we can use a method that computes the roots of a polynomial with coefficients given in a list. The roots can be real or complex numbers.Step:3 ：The coefficients of the equation are [3, -7, 27, -63].Step:4 ：The roots of the equation are approximately (0+3j), (0-3j), and (2.33).Step:5 ：The real part of the complex numbers is very close to zero, so we can consider them as pure imaginary numbers.Step:6 ：Final Answer: The roots of the equation are approximately (0+3j), (0-3j), and (boxed{2.33}).

Answer

Step: 1 ：We are given the cubic equation (3 x^{3}-7 x^{2}+27 x-63=0).Step: 2 ：To solve it, we can use a method that computes the roots of a polynomial with coefficients given in a list. The roots can be real or complex numbers.Step: 3 ：The coefficients of the equation are [3, -7, 27, -63].Step: 4 ：The roots of the equation are approximately (0+3j), (0-3j), and (2.33).Step: 5 ：The real part of the complex numbers is very close to zero, so we can consider them as pure imaginary numbers.Step: 6 ：Final Answer: The roots of the equation are approximately (0+3j), (0-3j), and (boxed{2.33}).

$3 x^{3}-7 x^{2}+27 x-63=0$

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