Let $a,b,$ and $c$ be the roots of the equation $2x^{3}-x^{2}+x+3=0$ . Find the value of the expression below.

$\large \frac{a^{3}-b^{3}}{a-b}+\frac{b^{3}-c^{3}}{b-c}+\frac{c^{3}-a^{3}}{c-a}$

The answer is -1.

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Using Vieta's formula ,

$a+b+c=\dfrac{1}{2} \\ ab+bc+ac = \dfrac{1}{2}$

Using Newton's identity ,

$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ac) = \left(\dfrac{1}{2}\right)^2 - 2\left(\dfrac{1}{2}\right) = \dfrac{-3}{4}$

By using identity $x^3-y^3=(x-y)(x^2+xy+y^2)$ ,the terms like $(a-b) , (b-c),(c-a)$ get cancelled $(a \neq b \neq c)$ and thus we can write the given expression as :

$a^2+ab+b^2+b^2+bc+c^2+c^2+ac+a^2 \\ = 2(a^2+b^2+c^2)+ab+bc+ac \\ = 2\left(\dfrac{-3}{4}\right) + \dfrac{1}{2} \\ =\Large\boxed{-1}$