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**Hint:**We start solving the by assuming the quadratic equation as ${{x}^{2}}+ax+b=0$. We use the ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+{{b}^{2}}+ab \right)$ for difference in cubes of roots and do subsequent calculations to find the value of sum of roots and product of roots. Using this sum and product, we find the quadratic equation.

**Complete step-by-step solution:**Given that we have the difference of the roots and difference between cubes of roots of the quadratic equation is 3 and 189. We need to check whether the quadratic equation ${{x}^{2}}\pm 9x+18=0$ or not.

Let us assume the quadratic equation be ${{x}^{2}}+ax+b=0$, and the roots of this quadratic equation be $\alpha $ and $\beta $. We know that sum of the roots $\alpha +\beta =-a$ and product of the roots $\alpha \beta =b$.

According to the problem, we have $\alpha -\beta =3$, and ${{\alpha }^{3}}-{{\beta }^{3}}=189$ -------(1).

We know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+{{b}^{2}}+ab \right)$. We use this result for equation (1).

We have got ${{\alpha }^{3}}-{{\beta }^{3}}=189$.

We have got $\left( \alpha -\beta \right)\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.

We substitute the value $\alpha -\beta =3$ now.

We have got $3\times \left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=189$.

We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=\dfrac{189}{3}$.

We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$ --------(2).

We have got ${{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +\alpha \beta +2\alpha \beta =63$ -------(3).

We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We use this in equation in (3).

So, we have got ${{\left( \alpha -\beta \right)}^{2}}+3\alpha \beta =63$.

We have got ${{3}^{2}}+3\alpha \beta =63$.

We have got $9+3\alpha \beta =63$.

We have got $3\alpha \beta =63-9$.

We have got $3\alpha \beta =54$.

We have got $\alpha \beta =\dfrac{54}{3}$.

We have got $\alpha \beta =18$ -------(4).

From equation (2), we get

We have got $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)=63$.

We have got ${{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta +\alpha \beta -\alpha \beta =63$.

We have got ${{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -\alpha \beta =63$ -------(5).

We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We use this in equation in (5).

We have got ${{\left( \alpha +\beta \right)}^{2}}-\alpha \beta =63$.

From equation (4), we use $\alpha \beta =18$.

We have got ${{\left( \alpha +\beta \right)}^{2}}-18=63$.

We have got ${{\left( \alpha +\beta \right)}^{2}}=63+18$.

We have got ${{\left( \alpha +\beta \right)}^{2}}=81$.

We have got $\alpha +\beta =\sqrt{81}$.

We have got $\alpha +\beta =\pm 9$ --------(6).

From equations (5) and (6), we have got $-a=\alpha +\beta =\pm 9$ and $b=\alpha \beta =18$.

Since $\alpha +\beta =\pm 9$, we can take the value of as $\pm 9$.

So, the quadratic equation ${{x}^{2}}\pm 9x+18=0$.

$\therefore$ The required quadratic equation is ${{x}^{2}}\pm 9x+18=0$.

**The correct option for the given problem is (a) True.**

**Note:**Here we can take the equation of the quadratic equation is $a{{x}^{2}}+bx+c=0$ and solve for the values of a, b and c by using the sum and product of the roots of the equation. But we need to solve for the value of ‘a’ again. Whenever we get to solve this type of problem, we start by assuming the appropriate quadratic equation.

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