91Ó°ÊÓ

We can find the roots of the given quadratic equation with the quadratic formula: \(r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Let's denote the roots as \(r_1\) and \(r_2\).

According to the condition, one root must be the square of the other. Let's denote the larger root as \(r_1\) and the smaller root as \(r_2\). So, we have \(r_1 = r_2^2\).

Vieta's formulas for a quadratic equation \(ax^2 + bx + c = 0\) are as follows: 1. Sum of roots, \(r_1 + r_2 = -\frac{b}{a}\) 2. Product of roots, \(r_1 \cdot r_2 = \frac{c}{a}\) Now, substitute the relationship \(r_1 = r_2^2\) into the Vieta's formulas: 1. \(r_2^2 + r_2 = -\frac{b}{a}\) 2. \(r_2^3 = \frac{c}{a}\)

We want to eliminate \(r_2\) from the two equations in Step 3, and then find an equation in terms of \(a\), \(b\), and \(c\) that matches one of the given options. 1. From the first equation, \(r_2 = -\frac{b}{a} - r_2^2\). Substitute the value of \(r_2^3\) from the second equation: \(r_2 = -\frac{b}{a} - \sqrt[3]{\frac{c}{a}}\). 2. Cube both sides of the equation: \(r_2^3 = (-\frac{b}{a}-\sqrt[3]{\frac{c}{a}})^3\). 3. Substitute the value of \(r_2^3\) from the second equation again: \(\frac{c}{a} = (-\frac{b}{a}-\sqrt[3]{\frac{c}{a}})^3\). 4. Expand the equation and simplify: \(c(a+b) = 3abc - a^3 - b^3 - bc(b+c)\). 5. Rearrange the terms: \(b^3 + ac(a+c) = 3abc - a^3 - bc(b+c)\). Comparing the resulting equation with the given options, we find that it matches option (b): \(b^{3}+a c(a+c)=3 a b c\) So, the correct answer is (b).