Question

How can $\left( {{x^3} - 1} \right)$ be factorized as where $\omega $ is one of the cube roots of unity?
A) $\left( {x - 1} \right)\left( {x - \omega } \right)\left( {x + {\omega ^2}} \right)$
B) $\left( {x - 1} \right)\left( {x - \omega } \right)\left( {x - {\omega ^2}} \right)$
C) $\left( {x - 1} \right)\left( {x + \omega } \right)\left( {x + {\omega ^2}} \right)$
D) $\left( {x - 1} \right)\left( {x + \omega } \right)\left( {x - {\omega ^2}} \right)$

Answer 1

Hint: First use the formula ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ to get the first root of the expression. After that use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to find the other two roots of the equation. Then subtract the terms from $x$ and multiply. The factorization is the desired result.

Complete step by step solution:
Factorization is the process of finding what to multiply together to get the given expression. It is like “splitting” an expression into a multiplication of simpler expressions. In the given expression we will take common and we will use the formula to factorize it further and by using the formula we can easily get the result, the formula is,
${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
Now for the given algebraic expression the factorizing is as follow,
$ \Rightarrow {x^3} - 1 = \left( {x - 1} \right)\left[ {{x^2} + x \times 1 + {1^2}} \right]$
Multiply the terms,
$ \Rightarrow {x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)$
So, the one factor is $\left( {x - 1} \right)$.
Now find the factors of the remaining part by the quadratic formula which is,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substitute, $a = 1$, $b = 1$ and $c = 1$ in the above formula,
\[ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 1 \times 1} }}{{2 \times 1}}\]
Multiply the terms,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 - 4} }}{2}$
Subtract the value inside the square root,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
We know that $i = \sqrt { - 1} $. Substitute it in the above equation,
$ \Rightarrow x = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$
So, the other 2 roots are $\dfrac{{ - 1 + i\sqrt 3 }}{2}$ and $\dfrac{{ - 1 - i\sqrt 3 }}{2}$.
Let us consider it $\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}$.
Now square the $\omega $,
\[ \Rightarrow {\omega ^2} = {\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)^2}\]
Square the term on the right side by the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$,
\[ \Rightarrow {\omega ^2} = \dfrac{{{{\left( { - 1} \right)}^2} + 2 \times - 1 \times i\sqrt 3 + {{\left( {i\sqrt 3 } \right)}^2}}}{4}\]
Simplify the terms,
\[ \Rightarrow {\omega ^2} = \dfrac{{1 - 2i\sqrt 3 + 3{i^2}}}{4}\]
Substitute the value ${i^2} = - 1$,
\[ \Rightarrow {\omega ^2} = \dfrac{{1 - 2i\sqrt 3 - 3}}{4}\]
Subtract the like terms in the numerator,
\[ \Rightarrow {\omega ^2} = \dfrac{{ - 2 - 2i\sqrt 3 }}{4}\]
Take 2 commons from the numerator and cancel out the common terms,
\[ \Rightarrow {\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}\]
So, the other two roots are $\omega ,{\omega ^2}$.
Thus, the factored form is $\left( {x - 1} \right)\left( {x - \omega } \right)\left( {x - {\omega ^2}} \right)$.

Hence, option (B) is correct.

Note: The number of zeroes of a polynomial is equal to the degree of the polynomial, and there is a well-defined mathematical relationship between the zeroes and the coefficients. Here, the number k is said to be a zero of the polynomial $p\left( x \right)$, if $p\left( k \right) = 0$.
There are various methods to find the zeros of the polynomial; we can do this with classical mathematics. We can also use calculus i.e. differentiation. All the methods will lead to the same answer.