Question

How do you factor completely \[{x^9} - 1\] ?

Answer 1

Hint: The given equation is given in complex form \[{x^9} - 1\] . we can convert this equation into cubes and factor accordingly with the cubic root formula. The key is in factoring the data as much as possible in simpler form. Factoring is the process of finding the factors and finding what to multiply together to get an expression. It is like "splitting" an expression into a multiplication of simpler expressions. To solve this equation, we apply simple methods.
Formula:
 \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]

Complete step by step solution:
We are given the polynomial equation can be converted into simpler form as follows:
 \[{x^9} - 1\] can be written as \[{({x^3})^{^3}} - {1^3}\] .
As both terms are perfect cubes, we can factor using the difference of cubes formula which is:
 \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]
Substituting the values, \[a = {x^3}\] and \[b = 1\] we get,
 \[ \Rightarrow {({x^3})^{^3}} - {1^3} = ({x^3} - {1^3})({({x^3})^2} + {x^3}1 + {1^2})\]
Now on RHS side, \[({x^3} - {1^3})\] can be further factored using the difference of cubes formulas as:
 \[ \Rightarrow ({x^3} - {1^3}) = (x - 1)({x^2} + x.1 + {1^2})\]
Therefore, we get the entire equation as follows:
 \[ \Rightarrow {({x^3})^{^3}} - {1^3} = (x - 1)({x^2} + x.1 + {1^2})({({x^3})^2} + {x^3}1 + {1^2})\]
Multiplying \[x\] with 1 and solving the power of 1,
 \[ \Rightarrow {({x^3})^{^3}} - 1 = (x - 1)({x^2} + x + 1)({({x^3})^2} + {x^3} + 1)\]
Finally, we multiply the exponents of \[{({x^3})^2}\] , we can get
 \[ \Rightarrow {({x^3})^{^3}} - 1 = (x - 1)({x^2} + x + 1)({x^6} + {x^3} + 1)\]
Hence, we arrive at the final factored solution as follows:
 \[ \Rightarrow {x^9} - 1 = (x - 1)({x^2} + x + 1)({x^6} + {x^3} + 1)\]
So, the correct answer is “\[ (x - 1)({x^2} + x + 1)({x^6} + {x^3} + 1)\]”.

Note: A polynomial is a mathematical equation made up of variables and coefficients that only includes the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
When you're given a pair of cubes to factor, use parentheses to keep track and simplify, especially the negative signs.
The key to remembering the difference of cubic formulas easily is to use “SOAP” (S-Same sign, O-Opposite, AP-Always positive) method as follows: \[{a^3} \pm {b^3} = (a\left[ S \right] b)({a^2}\left[ O \right] ab\left[ {AP} \right] {b^2})\]
Using factorization methods to solve further to get required solutions.