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Question: How do you express \({{\left( 1-i \right)}^{3}}\) in \(a+ib\) form?...

How do you express (1−i)3{{\left( 1-i \right)}^{3}} in a+iba+ib form?

Explanation

Solution

We first find the simplification of the given polynomial (1−i)3{{\left( 1-i \right)}^{3}} according to the identity (x−y)3=x3−3x2y+3xy2−y3{{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}. We need to simplify the cubic polynomial of difference of two terms. We replace it with x=1;y=ix=1;y=i. We also use i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.

Complete step-by-step solution:
We need to find the simplified form of (1−i)3{{\left( 1-i \right)}^{3}}.
We are going to use the identity (x−y)3=x3−3x2y+3xy2−y3{{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}.
We express (1−i)3{{\left( 1-i \right)}^{3}} as the cube of difference of two numbers. We take x=1;y=ix=1;y=i for the identity of (x−y)3=x3−3x2y+3xy2−y3{{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}.
(1−i)3=13−3×12×i+3×1×i2−i3{{\left( 1-i \right)}^{3}}={{1}^{3}}-3\times {{1}^{2}}\times i+3\times 1\times {{i}^{2}}-{{i}^{3}}
We have the relations for imaginary ii where i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
Therefore, the simplified form of (1−i)3{{\left( 1-i \right)}^{3}} is

& {{\left( 1-i \right)}^{3}} \\\ & =1-3i+3{{i}^{2}}-{{i}^{3}} \\\ & =1-3i-3+i \\\ & =-2-2i \\\ \end{aligned}$$ **Therefore, expressing ${{\left( 1-i \right)}^{3}}$ in $a+ib$ form, we get $$-2-2i$$.** **Note:** We also can use the binomial theorem to find the general form and then put the value of 3. We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of sum of two numbers. So, we put $n=3$. ${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. In this way we also simplify the term of ${{\left( a-b \right)}^{3}}$.