Question

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

Answer 1

Hint: We first find the simplification of the given polynomial ${{\left( 1-i \right)}^{3}}$ according to the identity ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$. We need to simplify the cubic polynomial of difference of two terms. We replace it with $x=1;y=i$. We also use ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.

Complete step-by-step solution:
We need to find the simplified form of ${{\left( 1-i \right)}^{3}}$.
We are going to use the identity ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$.
We express ${{\left( 1-i \right)}^{3}}$ as the cube of difference of two numbers. We take $x=1;y=i$ for the identity of ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$.
${{\left( 1-i \right)}^{3}}={{1}^{3}}-3\times {{1}^{2}}\times i+3\times 1\times {{i}^{2}}-{{i}^{3}}$
We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
Therefore, the simplified form of ${{\left( 1-i \right)}^{3}}$ is
\[\begin{align}
  & {{\left( 1-i \right)}^{3}} \\
 & =1-3i+3{{i}^{2}}-{{i}^{3}} \\
 & =1-3i-3+i \\
 & =-2-2i \\
\end{align}\]
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 ${{\left( a-b \right)}^{3}}$.