Question

How do you factor ${z^6} - 1$ ?

Answer 1

Hint: In this question, we are given an expression, and we have to factorize the given expression. It can be factorized in the following way:
We will write the expression as a difference between two squares. Then, we will use the formula of ${a^2} - {b^2}$ to find the factors of the given expression.
After that, we will get two factors. These two factors will have differences and sum between cubes. In order to solve these two factors, we will use the formula of ${a^3} - {b^3}$ and ${a^3} + {b^3}$ .

Formula used: 1) ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
2) ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
3) ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$

Complete step-by-step solution:
We are given an expression.
$ \Rightarrow {z^6} - 1$ …. (given)
We can also write this expression as difference of two squares–
$ \Rightarrow {\left( {{z^3}} \right)^2} - {\left( {{1^3}} \right)^2}$
Here, $a = {z^3}$ and $b = {1^3} = 1$ .
Now, we will expand the given expression using the formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ ,
$ \Rightarrow {\left( {{z^3}} \right)^2} - {\left( {{1^3}} \right)^2} = \left( {{z^3} + 1} \right)\left( {{z^3} - 1} \right)$
Now, we will expand both the factors using the formula of difference and sum of cubes.
The formulas to be used are –
$ \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$

Let us apply them in the factors above.
\[ \Rightarrow {\left( {{z^3}} \right)^2} - {\left( {{1^3}} \right)^2} = \left( {z + 1} \right)\left( {{z^2} - z + 1} \right)\left( {z - 1} \right)\left( {{z^2} + z + 1} \right)\]
Hence, these are the final factors of ${z^6} - 1$.

Note: Now, we will find the values of z using these factors. Make a note that the solutions will always be equal to the degree of the equation.
Values of z can be found by putting each factor equal to 0.
1) Putting $\left( {z + 1} \right) = 0$,
We get $z = - 1$.
2) Now, we will put $\left( {z - 1} \right) = 0$
We get $z = 1$.
3) Next, we will put ${z^2} - z + 1 = 0$
Using quadratic formula for this equation,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting all the values,
$ \Rightarrow x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1} }}{2}$
We get,
$ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - 4} }}{2}$
Let us subtract the term and we get
$ \Rightarrow x = \dfrac{{1 \pm \sqrt { - 3} }}{2}$
4) Now, we will put ${z^2} + z + 1 = 0$
Using quadratic formula for this equation,
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting all the values,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4 \times 1} }}{2}$
We get,
$ \Rightarrow x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
Hence, the six factors of ${z^6} - 1$ are $1$, $ - 1$, $\dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$ and $\dfrac{{1 \pm \sqrt { - 3} }}{2}$.
You can check that we got 6 factors.