Question

Factorize $ {x^3} - {1^3} $ .

Answer 1

Hint: To find the factors of $ {x^3} - {1^3} $ , we are going to use the Polynomial remainder theorem. According to this theorem, $ x - a $ is a factor of f(x) if $ f\left( a \right) = 0 $ . Here, $ f\left( 1 \right) = 0 $ , so $ \left( {x - 1} \right) $ is one of the factors of the given polynomial. To find the other factors, we will be using the long division method. Long division method is explained in detail below.

Complete step-by-step answer:
In this question, we have to factorize $ {x^3} - {1^3} $ .
Generally, polynomials can be factored using the Polynomial remainder theorem.
Let us see what this theorem is.
For a given polynomial, $ x - a $ is a divisor of f(x) if $ f\left( a \right) = 0 $ . That means if we are given a polynomial and if we put a in place of x in the given polynomial and if we get value 0, then $ x - a $ will be one of the factors of the given polynomial.
Here, in our given polynomial if let us find f(1).
 $ \Rightarrow f\left( 1 \right) = {\left( 1 \right)^3} - {1^3} = 1 - 1 = 0 $
Hence, $ \left( {x - 1} \right) $ is one of the factors of the given polynomial $ {x^3} - {1^3} $ .
Now, to find out other factors, we need to carry out a long division method.
Steps for long division:
First of all, write $ \left( {x - 1} \right) $ as divisor and $ {x^3} - {1^3} $ as dividend.
 $ \Rightarrow x - 1\left| \!{\overline {\,
 {{x^3} - {1^3}} \,}} \right. $
Now, we need $ {x^3} $ so multiply $ \left( {x - 1} \right) $ with $ {x^2} $ . Therefore, $ \left( {x - 1} \right){x^2} = {x^3} - {x^2} $
So, therefore now subtract $ {x^3} - {x^2} $ from $ {x^3} - {1^3} $
 $ \Rightarrow {x^3} - {1^3} - {x^3} + {x^2} = {x^2} - 1 $ and quotient $ = {x^2} $
Now, we need to get $ {x^2} $ , so multiply $ \left( {x - 1} \right) $ with x. Therefore, $ \left( {x - 1} \right)x = {x^2} - x $
So, therefore now subtract $ {x^2} - x $ from $ {x^2} - 1 $
 $ \Rightarrow {x^2} - 1 - {x^2} + x = x - 1 $ and quotient $ = {x^2} + x $
Now, we need to multiply $ \left( {x - 1} \right) $ with 1. Therefore, $ \left( {x - 1} \right)1 = x - 1 $
So, therefore now subtract $ x - 1 $ from $ x - 1 $
 $ \Rightarrow x - 1 - x + 1 = 0 $ and quotient $ = {x^2} + x + 1 $
Hence, the other factor of $ {x^3} - {1^3} $ is $ {x^2} + x + 1 $ .
Hence, $ {x^3} - {1^3} = \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: We can also solve this using the mathematical identity.
The identity is: $ {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) $
Here, $ a = x $ and $ b = 1 $ .
Therefore, $ {x^3} - {1^3} = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) $
Hence, the factors of $ {x^3} - {1^3} $ are $ \left( {x - 1} \right) $ and $ \left( {{x^2} + x + 1} \right) $ .