Question

How do you factor ${x^{16}} - 81$?

Answer 1

Hint: In this question, we are given a variable having power as $16$, and it is subtracting $81$ from it. This particular equation can be solved by different methods such as finding $16$ roots, or by difference of two squares methods.
Finding 16 roots looks like $(x - {x_0})(x - {x_1})(x - {x_2})...(x - {x_{15}})$
Whereas, difference between the two squares looks like $({x^8} - 9) \times ({x^8} + 9)$

Complete step-by-step solution:
1: When we solve ${x^{16}} - 81$, we find 16 roots.
$ \Rightarrow {x^{16}} = 81$
Shifting the powers,
$ \Rightarrow x = \dfrac{{81}}{{{1^{16}}}}$
Applying the factorisation formula which is,
${x_n} = {a^b}(\cos 2\pi .nb) + i\sin (2\pi .nb)$
Putting in the values,
\[ \Rightarrow {x_n} = {81^{\dfrac{1}{{16}}}}\left( {\cos \dfrac{{2\pi n}}{{16}}} \right) + i\sin \left( {\dfrac{{2\pi n}}{{16}}} \right)\]
Now that we have the above equation, with the numbers placed logically, we can divide the terms which are divisible. So, dividing the $16$ inside the brackets with the number $2$.
$ \Rightarrow {x_n} = {81^{\dfrac{1}{{16}}}}\left( {\cos \dfrac{{\pi n}}{8}} \right) + i\sin \left( {\dfrac{{\pi n}}{8}} \right)$
Also, we all know that $81$ is with the power $\dfrac{1}{{16}}$, which can be square rooted down,
So, square rooting the numbers,
$ \Rightarrow {x_n} = {3^{\dfrac{1}{4}}}\left( {\cos \dfrac{{\pi n}}{8}} \right) + i\sin \left( {\dfrac{{\pi n}}{8}} \right)$
These are the factors reached with method 1.
2: When we solve ${x^{16}} - 81$, we can do that as the difference of two squares.
$({x^8} - 9) \times ({x^8} + 9)$
Here, we can observe that,
${x^{16}}$ can also be written and considered as square of ${x^8} \times {x^8}$
$81$ can be written and considered as square of $9 \times 9$
According to mathematical formulas, when the binomials of two squares are multiplied as alternatively positive and negative terms, the middle term falls out.

Note: A binomial is a difference of squares if both terms are perfect squares.
$ \Rightarrow ({x^8} - 9) \times ({x^8} + 9) = {x^{16}} + 9{x^8} - 9{x^8} - 81$
We know that, $9{x^8} - 9{x^8} = 0$
So, keeping the equation in the brackets itself,
$ \Rightarrow ({x^8} - 9) \times ({x^8} + 9) = {x^{16}} - 81$
3: Difference of squares methods can be used even further like $({x^8} + 9)(({x^4} + 3)({x^4} - 3))$
Here, we can clearly see the difference of squares method which has been applied even more accurately,
So, for the equation ${x^{16}} - 81$
Using difference of squares,
$ \Rightarrow ({x^8} + 9) \times ({x^8} - 9)$
Now, using difference of squares on the second bracket of the above equation,
$ \Rightarrow ({x^8} + 9)(({x^4} + 3)({x^4} - 3))$
Therefore, for ${x^{16}} - 81$, the factors can be found in the above methods such as:
${x_n} = {3^{\dfrac{1}{4}}}\left( {\cos \dfrac{{\pi n}}{8}} \right) + i\sin \left( {\dfrac{{\pi n}}{8}} \right)$
$({x^8} - 9) \times ({x^8} + 9) = {x^{16}} - 81$
$({x^8} + 9)(({x^4} + 3)({x^4} - 3))$
Factors of a squared variable or a complex number can be found in many ways; some of them are given above and are mostly used. Difference of squares is one such method which solves an equation by forming two different simplified versions of the given function, factorising it I to smaller powered brackets.