Question

How do you factor \[{x^3} + 1\] ?

Answer 1

Hint: The given expression is simply we can say a perfect cube. So we will simply use the algebraic identities to expand it. And that expansion will be the factor form. Since the expression is having a degree tree there will be three factors of it. So let’s start!

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

Complete step by step solution:
Given that we have to factor \[{x^3} + 1\]
Here we are not getting any common terms nor any perfect term. But we can observe that these terms are simply cubes.
 So we can use the identity expansion of sum of two cubic terms as \[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\]
Here we have a=x and b=1.
 So just substituting them in the formula we get,
 \[ = {\left( x \right)^3} + {\left( 1 \right)^3}\]
On expanding we get,
 \[ = \left( {x + 1} \right)\left( {{x^2} - x \times 1 + {1^2}} \right)\]
Now any number or variable multiplied with 1 gives that number or variable itself. So multiplying x with 1 we get,
 \[ = \left( {x + 1} \right)\left( {{x^2} - x + {1^2}} \right)\]
Taking the square of 1 as any power of 1 is 1 only,
 \[ = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\]
These are the factors of the given expression. \[ = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\]
So, the correct answer is “ \[\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\] ”.

Note: Note that the answer so obtained is 100% correct. But if we go in depth since we said there should be three factors. We need to factorise the second bracket also using quadratic formula but this is optional only if the answer we have obtained is not present in the multiple choice.