### If $\alpha$ and $\beta$ are the zeros of polynomial $x^2-x-6,$ find a polynomial whose zeros are $\dfrac{ \alpha^2 }{ \beta^2 }$ and $\dfrac{ \beta^2 }{ \alpha^2 }.$

$k \left( x^2 - \dfrac{ 97 } { 36 }x + 1 \right)$

Step by Step Explanation:
1. On comparing the polynomial $x^2-x-6,$ with the standard form $ax^2 + bx + c = 0$, we get:
$$a = 1 \\ b = -1 \\ c = -6$$
2. Since, sum of zeros $\alpha + \beta = \dfrac { -b } { a } = \dfrac { 1 } { 1 } = 1$
Product of zeros $\alpha \beta = \dfrac { c } { a } = \dfrac { -6 }{ 1 } = -6$
3. Let $S$ and $P$ respectively be the sum and products of zeros of the required polynomial.
4. $S = \dfrac { \alpha^2 } { \beta^2 } + \dfrac{ \beta^2 } { \alpha^2 } = \dfrac {\alpha^4 + \beta^4 } { \alpha^2 \beta^2} = \dfrac{(\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 } { (\alpha \beta)^2} = \dfrac{ \Big((\alpha + \beta)^2 - 2\alpha\beta \Big)^2 - 2(\alpha\beta)^2 } { (\alpha \beta)^2 }$
5. $S = \dfrac{\Big((1)^2 - 2(-6)\Big)^2 - 2(-6)^2 } { (-6)^2 } = \dfrac{ 97 } { 36 }$
6. $P = \bigg( \dfrac { \alpha ^2 } { \beta ^2 } \bigg) \bigg( \dfrac { \beta ^2 } { \alpha ^2 } \bigg) = 1$
7. Thus, required polynomial will be \begin{align} & k \left( x^2 - Sx + P \right) = k \left( x^2 - \dfrac{ 97 } { 36 }x + 1 \right) && [\text{ Where k is a constant }] \end{align}