Question

X
${x}^{3} - 3i {x}^{2} - (9 + 6i)x + 2 + 17i = 0$
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Answer 1

Ответ:

x1=i-(-1)^(1/6)*(1+i)*(1-i3^(1/2))/2^(1/3)+(-1)^(5/6)*(1+i3^(1/2))/2^

(2/3);

x2=i+(-1)^(5/6)*(1-i3^91/2))/2^(2/3)-1^(1/6)*(1+i)*(1+i3^(1/2))/2^(1/3);

x3=i-(-1)^(5/6)*2^(1/3)+(1+i)*(-1)^(1/6)*2^(2/3);

Объяснение:

x^(3)-3ix^(2)-(9+6i)x+2+17i=0

x^(3)-3ix^(2)-(9+6i)x+(2+17i)=0

x^(3)-3ix^(2)-(9+6i)x=-2-17i {приведем кубическое уравнение в стандартный вид}  

(x-i)^(3)-(6+6i)(x-i)+(8i+10)=0

x^(3)+i(-3x^(2)-6x+17)-9x+2=0 {подставим в формулу для нахождение кубов}

x1=i-(-1)^(1/6)*(1+i)*(1-i3^(1/2))/2^(1/3)+(-1)^(5/6)*(1+i3^(1/2))/2^

(2/3);

x2=i+(-1)^(5/6)*(1-i3^91/2))/2^(2/3)-1^(1/6)*(1+i)*(1+i3^(1/2))/2^(1/3);

x3=i-(-1)^(5/6)*2^(1/3)+(1+i)*(-1)^(1/6)*2^(2/3);