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

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(C) Given the equation $x^3 + 8x + 1 = 0$,since $a, b, c$ are roots,we have $x^3 = -(8x + 1)$.Thus,$8x + 1 = -x^3$.Substituting this into the expressi

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$-16$

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(C) Given the equation $x^3 + 8x + 1 = 0$,since $a, b, c$ are roots,we have $x^3 = -(8x + 1)$.Thus,$8x + 1 = -x^3$.Substituting this into the expression:$\frac{bc}{(-b^3)(-c^3)} + \frac{ac}{(-a^3)(-c^3)} + \frac{ab}{(-a^3)(-b^3)}$$= \frac{bc}{b^3c^3} + \frac{ac}{a^3c^3} + \frac{ab}{a^3b^3}$$= \frac{1}{b^2c^2} + \frac{1}{a^2c^2} + \frac{1}{a^2b^2}$$= \frac{a^2 + b^2 + c^2}{a^2b^2c^2}$Using Vieta's formulas for $x^3 + 0x^2 + 8x + 1 = 0$:Sum of roots $\Sigma a = a + b + c = 0$.Sum of roots taken two at a time $\Sigma ab = ab + bc + ca = 8$.Product of roots $abc = -1$.Now,$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0^2 - 2(8) = -16$.Also,$(abc)^2 = (-1)^2 = 1$.Therefore,the value is $\frac{-16}{1} = -16$.

Let $a$,$b$,and $c$ 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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