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# The number $\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}$ is equal to:

Last updated date: 29th Feb 2024
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Hint:To find the value of the term we use the complex number multiplication expressed in form of the complex number as $a+ib$ where the variable a and b are real numbers with the value of $i$ as imaginary unit, when multiplying the value of $i\times i$ we get the result of product as $-1$ which can be used in the above question.

Complete step by step solution:
Now as given in the question, the term $\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}$ first need to simplify it in terms of $i$ and constant numbers. Expanding the numerator and denominator in terms of simpler complex numbers, we get the numerator as:
$\Rightarrow {{\left( 1-i \right)}^{3}}=1-i-3i+3{{i}^{2}}$
And the denominator is written as:
$\Rightarrow 1-{{i}^{3}}=1+i$
Now placing the expanding part of the numerator and the denominator we get the term as:
$\Rightarrow \dfrac{1-\left( -i \right)-3i+3{{i}^{2}}}{1+i}$
$\Rightarrow \dfrac{-2-2i}{1+i}$
$\Rightarrow \dfrac{-2\left( 1+i \right)}{1+i}$
$\Rightarrow -2$
Therefore, the value of the term $\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}$ is $-2$.

Note: Another method to solve the question is by:
$\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)}{1-1\times -i}$
$\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)}{1+i}$
$\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{-2\left( 1+i \right)}{1+i}$
$\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=-2$