Question

Solve it.: (-2-1/3 iota)^{3}.

Answer 1

Solution:

Given:

$\implies \bigg(-2-\dfrac{1}{3}\iota \bigg)^{3}$

$\sf{\implies \Bigg[-\bigg(2 + \dfrac{1}{3}\iota \bigg)\Bigg]^{3}}$

$\sf{\implies -\bigg(2+\dfrac{1}{3}\iota \bigg)^{3}}$

By using (a + b)³ = a³ + b³ + 3ab(a + b)

$\sf{\implies -\Bigg[(2)^{3}+\bigg(\dfrac{\iota}{3}\bigg)^{3}+3(2)\bigg(\dfrac{\iota}{3}\bigg)\bigg(2+\dfrac{\iota}{3}\bigg)\Bigg]}$

$\sf{\implies - \Bigg[8+\dfrac{\iota^{3}}{27}+4\iota + \dfrac{2\iota^{2}}{3}\Bigg]}$

$\sf{\implies -\Bigg[8-\dfrac{\iota}{27}+4\iota - \dfrac{2}{3}\Bigg]}$

$\sf{\implies - \Bigg[\bigg(8-\dfrac{2}{3}\bigg) - \iota \bigg(\dfrac{1}{27}-4\bigg)\Bigg]}$

$\implies -\Bigg[\dfrac{22}{3}- \iota\bigg(\dfrac{-107}{27}\bigg)\Bigg]$

${\boxed{\boxed{\sf{\implies - \dfrac{22}{3}-\dfrac{107}{27}\iota}}}}$