Question

Factorise the expression : ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$.

Answer 1

Hint: As we are given in the question factorise ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$. ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$ is an algebraic expression which we need to factorise. The process of finding two or more expressions whose product is the given expression is called the factorization of algebraic expressions. We can factorise ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$ by using different identities.
Formulae used:
1. ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
2. ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
3. ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
4. \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step-by-step solution:
Given: Factorise ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$
As we can see ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$ looks familiar to ${a^3} + {b^3}$. And its identity is ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
So, we can expand ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$ using the above written identity as,
Here, we take $a = x + 1$ and $b = x - 1$
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {x + 1 + x - 1} \right)\left( {{{\left( {x + 1} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right) + {{\left( {x - 1} \right)}^2}} \right)$
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {x + x} \right)\left( {{{\left( {x + 1} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right) + {{\left( {x - 1} \right)}^2}} \right)$
On addition of like terms, we get
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{{\left( {x + 1} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right) + {{\left( {x - 1} \right)}^2}} \right)$
Here, we can see that our expression is now in the form of different identities.
We can use ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ identity to expand ${\left( {x + 1} \right)^2}$.
Similarly, we can use ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ identity to expand ${\left( {x - 1} \right)^2}$.
And also, here we have the expansion of \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]. So we can write $\left( {x + 1} \right)\left( {x - 1} \right) = {x^2} - {1^2}$
On expansion using these identities, we get
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{x^2} + {1^2} + 2x - \left( {{x^2} - {1^2}} \right) + {x^2} + {1^2} - 2x} \right)$
On simplifying, we get
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{x^2} + {1^2} + 2x - {x^2} + {1^2} + {x^2} + {1^2} - 2x} \right)$
On addition and subtraction of like terms, we get
(Like terms are the terms that have the same variables and powers, unlike terms whose variables and powers are different from each other. The coefficients do not need to match)
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{x^2} + {1^2} + {1^2} + {1^2}} \right)$
Square of $1$ is $1$, therefore we get
$ \Rightarrow {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{x^2} + 1 + 1 + 1} \right)$
On addition, we get
$\therefore {\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = \left( {2x} \right)\left( {{x^2} + 3} \right)$
Therefore, factors of ${\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}$ are $\left( {2x} \right)\left( {{x^2} + 3} \right)$.

Note: In the question, we used different identities to factorise the algebraic expression. Algebraic expressions can be factored using many methods: factorization using common factors, factorization using identities, and factorization by regrouping terms. While solving algebraic expression types of questions, be careful about identities and be careful about how to add or multiply the exponents.