Question

If ${\text{x = 1 + i}}$, then what is the value of ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1$?
A) $6{\text{i - 3}}$
B) ${\text{ - }}6{\text{i + 3}}$
C) ${\text{ - }}6{\text{i - 3}}$
D) $6{\text{i + 3}}$

Answer 1

In this question we have to find the value of given expansion. For that we are going to solve this using binomial expansion.
Here we used given to solve this problem, that implies the value of ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1$ and by using binomial expansion, we expansion the value of ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1$.
Here binomial expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${\left( {{\text{x + y}}} \right)^{\text{n}}}$ into a sum involving terms of the form ${\text{a}}{{\text{x}}^{\text{b}}}{{\text{y}}^{\text{c}}}$, where the exponents ${\text{b}}$ and ${\text{c}}$ are nonnegative integers with ${\text{b + c = n}}$, and the coefficients a of each term is a specific positive integer depending on ${\text{n}}$and ${\text{b}}$.

Formula used: ${{\text{i}}^0}{\text{ = 1}}$
${{\text{i}}^{\text{1}}}{\text{ = i}}$
${{\text{i}}^{\text{2}}}{\text{ = - 1}}$

Complete step-by-step answer:
An imaginary number ${\text{bi}}$ can be added to a real number ${\text{a}}$ to a complex number of the form ${\text{a + bi}}$.
Given that ${\text{x = 1 + i}}$
To find the value of ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1$.
Now, ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1$ ${\text{ = }}{\left( {1 + {\text{i}}} \right)^6} + {\left( {1 + {\text{i}}} \right)^4} + {\left( {1 + {\text{i}}} \right)^2} + 1$
[Here we substitute the value of ${\text{x = 1 + i}}$]
$\Rightarrow \left( {1{\text{ }} + {\text{ }}6{\text{i - 15 - 20i + 15 + 6i - 1}}} \right) + {\left( {1 + {\text{i}}} \right)^4} + {\left( {1 + {\text{i}}} \right)^2} + 1$
By using binomial theorem, ${\left( {{\text{a + b}}} \right)^{\text{n}}} = \left( {{\text{n}}{{\text{C}}_0}} \right){{\text{a}}^{\text{n}}}{{\text{b}}^0} + \left( {{\text{n}}{{\text{C}}_1}} \right){{\text{a}}^{{\text{n - 1}}}}{{\text{b}}^1} + ... + \left( {{\text{n}}{{\text{C}}_{\text{n}}}} \right){{\text{a}}^0}{{\text{b}}^{\text{n}}}$,
We can expand the first term,
$\Rightarrow {\text{ - 8i + }}{\left( {1 + {\text{i}}} \right)^4} + {\left( {1 + {\text{i}}} \right)^2} + 1$
Using binomial expansion,
$\Rightarrow {\text{ - 8i + }}\left( {1 + 4{\text{i - 6 - 4i + 1}}} \right) + {\left( {1 + {\text{i}}} \right)^2} + 1$
By using binomial theorem, ${\left( {{\text{a + b}}} \right)^{\text{n}}} = \left( {{\text{n}}{{\text{C}}_0}} \right){{\text{a}}^{\text{n}}}{{\text{b}}^0} + \left( {{\text{n}}{{\text{C}}_1}} \right){{\text{a}}^{{\text{n - 1}}}}{{\text{b}}^1} + ... + \left( {{\text{n}}{{\text{C}}_{\text{n}}}} \right){{\text{a}}^0}{{\text{b}}^{\text{n}}}$,
We expand the second term,
$\Rightarrow {\text{ - 8i - 4 + }}{\left( {1 + {\text{i}}} \right)^2} + 1$
Using binomial expansion,
$\Rightarrow {\text{ - 8i - 4 + }}\left( {{1^2} + {\text{ }}2 \times 1{\text{i + }}{{\text{i}}^2}} \right) + 1$
By using binomial theorem, ${\left( {{\text{a + b}}} \right)^{\text{n}}} = \left( {{\text{n}}{{\text{C}}_0}} \right){{\text{a}}^{\text{n}}}{{\text{b}}^0} + \left( {{\text{n}}{{\text{C}}_1}} \right){{\text{a}}^{{\text{n - 1}}}}{{\text{b}}^1} + ... + \left( {{\text{n}}{{\text{C}}_{\text{n}}}} \right){{\text{a}}^0}{{\text{b}}^{\text{n}}}$,
We expand the third term,
$\Rightarrow {\text{ - 8i - 4 + 2i + 1}}$
By adding the terms, we get
$\Rightarrow - 3 - 6{\text{i}}$
Hence, the value of ${{\text{x}}^6} + {\text{ }}{{\text{x}}^4} + {\text{ }}{{\text{x}}^2} + {\text{ }}1 = {\text{ - 3 - 6i}}$

$\therefore$ The option C is the correct answer.

Note: A binomial is a polynomial with two variables. It describes the algebraic expansion of the powers. The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas of mathematics.