Find the value of ${(1 + i)}^{3} + {(1 - i)}^{6}$ .

Open in App

Solution

To find : $(1 + i)^{3} + (1 - i)^{6}$

$(1 - i)^{6} = ((1 - i)^{2})^{3}$

$= (1^{2} + i^{2} - 2 i)^{3}$ $[i^{2} = - 1]$

$= (1 - 1 - 2 i)^{3} = - 8 i^{3} = + 8 i$ _ (1)

$(1 + i)^{3} = 1^{3} + i^{3} + 3 i + 3 i^{2}$

$= 1 - i + 3 i - 3$

$= 2 i - 2$ _ (2)

Add eq (1) and (2)

$(1 + i)^{3} + (1 - i)^{6} = 2 i - 2 + 8 i$

$(1 + i)^{3} + (1 - i)^{6} = 10 i - 2$

Suggest Corrections

Similar questions

(1 + i)^{6} + (1 - i)^{6}

(1 + i)^{6} + (1 - i)^{3}

\frac{i^{6} + i^{7} + i^{9}}{i^{2} + i^{3}}

\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}}

I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x d x

Join BYJU'S Learning Program