Let a, b, c be the roots of the equation x^3 | vedclass

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(C) Given $x^3 + 8x + 1 = 0$,the roots are $a, b, c$. Thus,$x^3 = -(8x + 1)$.For any root $r$,$r^3 = -(8r + 1)$,which implies $8r + 1 = -r^3$.Substitu

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$-16$

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(C) Given $x^3 + 8x + 1 = 0$,the roots are $a, b, c$. Thus,$x^3 = -(8x + 1)$.For any root $r$,$r^3 = -(8r + 1)$,which implies $8r + 1 = -r^3$.Substituting this into the expression:$\frac{bc}{(-b^3)(-c^3)} + \frac{ac}{(-a^3)(-c^3)} + \frac{ab}{(-a^3)(-b^3)}$$= \frac{bc}{b^3 c^3} + \frac{ac}{a^3 c^3} + \frac{ab}{a^3 b^3} = \frac{1}{b^2 c^2} + \frac{1}{a^2 c^2} + \frac{1}{a^2 b^2}$$= \frac{a^2 + b^2 + c^2}{a^2 b^2 c^2}$From the equation $x^3 + 0x^2 + 8x + 1 = 0$,we have $\sum a = 0$,$\sum ab = 8$,and $abc = -1$.$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = (0)^2 - 2(8) = -16$.$(abc)^2 = (-1)^2 = 1$.Therefore,the value is $\frac{-16}{1} = -16$.

Let $a, b, c$ be the roots of the equation $x^3 + 8x + 1 = 0$. Then the value of $\frac{bc}{(8b + 1)(8c + 1)} + \frac{ac}{(8a + 1)(8c + 1)} + \frac{ab}{(8a + 1)(8b + 1)}$ is equal to

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