If the equations ax^2 + bx + a = 0 and x^3 - | vedclass

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(C) Given the cubic equation $x^3 - 2x^2 + 2x - 1 = 0$.Factoring,we get $(x - 1)(x^2 - x + 1) = 0$.The roots are $x = 1$ and $x = \frac{1 \pm i\sqrt{3

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(C) Given the cubic equation $x^3 - 2x^2 + 2x - 1 = 0$.Factoring,we get $(x - 1)(x^2 - x + 1) = 0$.The roots are $x = 1$ and $x = \frac{1 \pm i\sqrt{3}}{2} = -\omega^2, -\omega$ (where $\omega$ is the complex cube root of unity).Since the quadratic equation $ax^2 + bx + a = 0$ has two common roots with the cubic equation,these roots must be the complex roots $-\omega$ and $-\omega^2$.For a quadratic equation $ax^2 + bx + a = 0$,the product of roots is $\frac{a}{a} = 1$.The product of the common roots is $(-\omega)(-\omega^2) = \omega^3 = 1$,which is consistent.The sum of the roots is $-\frac{b}{a} = -\omega - \omega^2 = -(\omega + \omega^2) = -(-1) = 1$.Thus,$-\frac{b}{a} = 1 \implies b = -a$,which means $a + b = 0$.

If the equations $ax^2 + bx + a = 0$ and $x^3 - 2x^2 + 2x - 1 = 0$ have two common roots,then what is the value of $a + b$?

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