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(A) Given equation is $x^3-3x^2+3x-9=0$. Factoring the expression,we get $x^2(x-3) + 3(x-3) = 0$. $(x-3)(x^2+3) = 0$. Thus,one root is $x = 3$. For th

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$3, 1+2\omega, 1+2\omega^2$

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(A) Given equation is $x^3-3x^2+3x-9=0$. Factoring the expression,we get $x^2(x-3) + 3(x-3) = 0$. $(x-3)(x^2+3) = 0$. Thus,one root is $x = 3$. For the other roots,we solve $x^2+3=0$,which implies $x^2 = -3$. Since $\omega^2+\omega+1=0$,we know $\omega^2+\omega = -1$. Testing $x = 1+2\omega$: $(1+2\omega)^2 + 3 = 1 + 4\omega + 4\omega^2 + 3 = 4(1+\omega+\omega^2) = 0$. Similarly,$x = 1+2\omega^2$ also satisfies the equation because $(1+2\omega^2)^2 + 3 = 1 + 4\omega^2 + 4\omega^4 + 3 = 1 + 4\omega^2 + 4\omega + 3 = 4(1+\omega+\omega^2) = 0$. Therefore,the roots are $3, 1+2\omega, 1+2\omega^2$.

Roots of the equation $x^3-3x^2+3x-9=0$ are ......

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