Question

Simplify the following :
$\dfrac{{{{\left( {1 - i} \right)}^3}}}{{1 - {i^3}}}$

Answer 1

Hint:- As we know $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$.

Complete step-by-step answer:
We can write the given equation as $\dfrac{{{{\left( {1 - i} \right)}^3}}}{{{1^3} - {i^3}}}$
And as we know $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
Then we can write the equation as $\dfrac{{{{\left( {1 - i} \right)}^3}}}{{\left( {1 - i} \right)\left( {{1^2} + {i^2} + 1i} \right)}}$
$= \dfrac{{{{\left( {1 - i} \right)}^2}}}{{\left( {{1^2} + {i^2} + 1i} \right)}}$ as we know $\left( {{i^2} = - 1} \right)$

$= \dfrac{{\left( {{1^2} - 2i + {i^2}} \right)}}{i} \\\ = \dfrac{{ - 2i}}{i} \\\ = - 2 \\\$

Note:- In such a type of question, First apply algebraic formula to reduce it. And then solve to get the result.