Question

How do you factor completely $x^3-1?$

Answer 1

Hint:
To find out factors of the above question, we will solve the cubes of the given terms with the help of identities and then put the given values into the identities. In the first method, we will use the identity in which cubes of two numbers are subtracted and in the second method, we use the identity in which cubes of two numbers are added.

Formula used:
$a^3-b^3=(a-b)(a^2+ab+b^2)$

Complete step by step solution:
We can factor $x^3-1$ by using formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ .
Here we will consider $a=x$
And $b=1$ as we know that $1^3=1$
Put these values into the formula.
$x^3-(1{)}^3=(x+(-1{)}^3)(x^2+x(1)+(1{)}^2)$
Simplify above equation
$\Rightarrow (x-1)(x^2+x+1)$

Hence, $(x-1)$ and $(x^2+x+1)$ are two factors of $x^3-1$.

Note:
Second method to solve above question:
Formula used:
$a^3+b^3=(a+b)(a^2-ab+b^2)$
Complete step by step solution:
We can factor $x^3-1$ by using formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ .
Here we will consider $a=x$
And $b=-1$ as we know that $(-1{)}^3=-1$
Put these values into a formula.
$x^3+(-1{)}^3=(x+(-1))(x^2-x(-1)+(-1{)}^2)$
Simplify above equation
$\Rightarrow (x-1)(x^2+x+1)$
Hence $(x-1)$ and $(x^2+x+1)$ are two factors of $x^3-1$ .