Question:
$ {{(1+i)}^{3}}+{{(1-i)}^{3}} $
$ {{(1+i)}^{3}}+{{(1-i)}^{3}} $
Updated On: Jun 14, 2022
- $ 1 $
- $ -2 $
- $ 0 $
- $ -4 $
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The Correct Option is D
Solution and Explanation
$ {{(1+i)}^{3}}+{{(1-i)}^{3}}={{(1+i)}^{2}}(1+i)+{{(1-i)}^{2}}(1-i) $ $ =(1+{{i}^{2}}+2i)(1+i)+(1+{{i}^{2}}-2i)\,(1-i) $ $ \left[ \begin{align} & \because \,\,\,\,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\ & {{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ \end{align} \right] $ $ =2i(1+i)+(-2i)(1-i) $ $ (\because \,\,{{i}^{2}}=-1) $ $ =2i+2{{i}^{2}}-2i+2{{i}^{2}} $ $ =4{{i}^{2}}=-4 $ $ (\because \,\,{{i}^{2}}=-1) $
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
