Cubic function

From Cardano's method , we have

$x_1$ (the real root of the cubic equation) = $S+T-\frac{b}{3a}$

$\implies x_1+\frac{b}{3a}=S+T$

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We know that $S=(R+\sqrt{Q^3+R^2})^{\frac{1}{3}}$ and $T=(R-\sqrt{Q^3+R^2})^{\frac{1}{3}}$

Where $R=\dfrac{9abc-27a^2d-2b^3}{54a^3}$ and $Q=\dfrac{3ac-b^2}{9a^2}$

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$(x_1+\frac{b}{3a})^3=((R+\sqrt{Q^3+R^2})^{\frac{1}{3}}+(R-\sqrt{Q^3+R^2})^{\frac{1}{3}})^3$

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We have $(m+n)^3=m^3+n^3+3mn(m+n)$

Hence, using the above identity and expanding we obtain the following expression.

$(x_1+\frac{b}{3a})^3=2R-3Q(x_1+\frac{b}{3a})$

From given data, we deduce that $R=\dfrac{65}{54}$ and $Q=\dfrac{-4}{9}$

Now, we can easily verify the options, fetching us the answer $(x_1+\frac{b}{3a})=\boxed{\dfrac{5}{3}}$

From cardano's formula ,we have,

$x_1=S+T-\frac{b}{3a}$ ,

Where $x_1$ is a real root.

$\implies x_1+\frac{b}{3a}=S+T$

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We know that from cardano's formula, $S=(R+\sqrt{Q^3+R^2})^{\frac{1}{3}}$ and $T=(R-\sqrt{Q^3+R^2})^{\frac{1}{3}}$

Where $R=\dfrac{9abc-27a^2d-2b^3}{54a^3}$ and $Q=\dfrac{3ac-b^2}{9a^2}$

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By given data, $R=\dfrac{65}{54}$ and $Q=\dfrac{-4}{9}$

substituting R and Q to get S and T,

$S=\dfrac{4}{3}$ and $T=\dfrac{1}{3}$

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$\implies x_1+\frac{b}{3a}=\frac{4}{3}+\frac{1}{3}$

$\implies \boxed{x_1+\dfrac{b}{3a}=\dfrac{5}{3}}$