If a, b, c are distinct real numbers and a^3 | vedclass

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(B) Given the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.This can be written as $\frac{1}{2}(a + b + c)[(a -

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$\frac{c}{a}$

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(B) Given the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.This can be written as $\frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0$.Since $a, b, c$ are distinct,$(a - b)^2 + (b - c)^2 + (c - a)^2 
eq 0$.Therefore,we must have $a + b + c = 0$.For the quadratic equation $ax^2 + bx + c = 0$,if we substitute $x = 1$,we get $a(1)^2 + b(1) + c = a + b + c$.Since $a + b + c = 0$,$x = 1$ is one root of the equation.Let the roots be $\alpha$ and $\beta$. We know $\alpha \cdot \beta = \frac{c}{a}$.Since $\alpha = 1$,the other root $\beta = \frac{c}{a}$.

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

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