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According to the question we need to find the value of the square of $ - \dfrac{{13}}{{17}}$. So, to find the value of square of $ - \dfrac{{13}}{{17}}$

Let us assume that $n = - \dfrac{{13}}{{17}}$ ,

Then to find the square of $ - \dfrac{{13}}{{17}}$ , we need to find the value of ${n^2}$ with $n = - \dfrac{{13}}{{17}}$.

Then, ${n^2} = n \times n$will become:

${n^2} = n \times n$

Now, putting $n = - \dfrac{{13}}{{17}}$we will get ${n^2}$ as:

$

\Rightarrow {\left( { - \dfrac{{13}}{{17}}} \right)^2} = \left( { - \dfrac{{13}}{{17}}} \right) \times \left( { - \dfrac{{13}}{{17}}} \right) \\

\Rightarrow {\left( { - \dfrac{{13}}{{17}}} \right)^2} = \dfrac{{169}}{{289}} \\

$

That is, the square of $ - \dfrac{{13}}{{17}}$will be $\dfrac{{169}}{{289}}$ a positive real number. Since as per the algebraic laws of multiplication a negative real number when multiplied with another negative real number gives a positive real number since:

$\left( - \right) \times \left( - \right) = \left( + \right)$

Thus, the sign of the product of a minus with a minus will always be a plus sign.

Therefore, the sign of $\dfrac{{169}}{{289}}$will be plus.

Therefore, Square of $ - \dfrac{{13}}{{17}}$ is ………$\dfrac{{169}}{{289}}$…………..

Note: The cube of the real number is calculated as the number multiplied with itself thrice. If the number is $n,$ then ${n^3} = n \times n \times n$ . The cube of a negative real number will always be negative because :

$

{\left( { - n} \right)^3} = \left( { - n} \right) \times \left( { - n} \right) \times \left( { - n} \right) \\

\Rightarrow {\left( { - n} \right)^3} = \left( {{n^2}} \right)\left( { - n} \right) \\

\Rightarrow {\left( { - n} \right)^3} = - {\left( n \right)^3} \\

$

Also, for a positive real number we do not need to put the plus sign additionally, because no sign before a real number conventionally means the number is positive.