Question

How do you factor $81{x^4} - 1$?

Answer 1

Hint: The algebraic expression given in the question can be solved with the help of algebraic identities. Let us first know what an algebraic identity is. It is an algebraic equation which holds true for all the values of its variable. There are four important algebraic identities as given below:
1. ${(a + b)^2} = {a^2} + 2ab + {b^2}$
2. ${(a - b)^2} = {a^2} - 2ab + {b^2}$
3. ${a^2} - {b^2} = (a + b)(a - b)$
4. $(x + a)(x + b) = {x^2} + x(a + b) + ab$
Here to solve the given question, we will use the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$.

Complete Step by Step Solution:
The given algebraic expression is $81{x^4} - 1$. It is in the form of ${a^2} - {b^2}$ in the identity ${a^2} - {b^2} = (a + b)(a - b)$. So now we will find out $a$ and $b$, such that
For $a$:
$ \Rightarrow {a^2} = 81{x^4}$
We know that $81 = {9^2}$ and ${x^4} = {({x^2})^2}$. On using these two expressions to simplify $81{x^4}$, we will get
$ \Rightarrow {a^2} = ({9^2}){({x^2})^2}$
When we square root both the sides of the equation, we will get
$ \Rightarrow \sqrt {{a^2}} = \sqrt {({9^2}){{({x^2})}^2}} $
It must be known that for any number $n$, $\sqrt {{{(n)}^2}} = n$. On applying this further, we will get
$ \Rightarrow a = 9{x^2}$
Similarly for $b$:
$ \Rightarrow {b^2} = 1$
When we square root both the sides of the equation, we will get
$ \Rightarrow \sqrt {{b^2}} = \sqrt 1 $
It must be known that $\sqrt 1 = \pm 1$ but we will neglect the negative value. So we will get
$ \Rightarrow b = 1$
Hence, we have $a = 9{x^2}$ and $b = 1$. On substituting these values on the right hand side of the identity ${a^2} - {b^2} = (a + b)(a - b)$, we will get
$ \Rightarrow {a^2} - {b^2} = (a + b)(a - b)$
$ \Rightarrow 81{x^4} - 1 = (9{x^2} + 1)(9{x^2} - 1)$ {Let it be equation (1)}
On factorising $81{x^4} - 1$, we got $(9{x^2} + 1)(9{x^2} - 1)$.
But we can further factorise $9{x^2} - 1$ using the same identity ${a^2} - {b^2} = (a + b)(a - b)$. Therefore we will find $a$ and $b$ again with respect to $9{x^2} - 1$.
For $a$:
$ \Rightarrow {a^2} = 9{x^2}$
We know that $9 = {3^2}$ and ${x^2} = {(x)^2}$. On using these two expressions to simplify $9{x^2}$, we will get
$ \Rightarrow {a^2} = ({3^2})({x^2})$
When we square root both the sides of the equation, we will get
$ \Rightarrow \sqrt {{a^2}} = \sqrt {({3^2})({x^2})} $
It must be known that for any number $n$, $\sqrt {{{(n)}^2}} = n$. On applying this further, we will get
$ \Rightarrow a = 3x$
Similarly for $b$:
$ \Rightarrow {b^2} = 1$
When we square root both the sides of the equation, we will get
$ \Rightarrow \sqrt {{b^2}} = \sqrt 1 $
It must be known that $\sqrt 1 = \pm 1$ but we will neglect the negative value. So we will get
$ \Rightarrow b = 1$
Hence, we have $a = 3x$ and $b = 1$. On substituting these values on the right hand side of the identity ${a^2} - {b^2} = (a + b)(a - b)$, we will get
$
   \Rightarrow {a^2} - {b^2} = (a + b)(a - b) \\
   \Rightarrow 9{x^2} - 1 = (3x + 1)(3x - 1) \\
 $
On factorising $9{x^2} - 1$, we got $(3x + 1)(3x - 1)$. Now substituting the obtained factors of $9{x^2} - 1$ in equation (1), we get
 $ \Rightarrow 81{x^4} - 1 = (9{x^2} + 1)(3x + 1)(3x - 1)$

Hence, the factors of $81{x^4} - 1$ are $(9{x^2} + 1)$, $(3x + 1)$ and $(3x - 1)$.

Note:
Two of the mentioned identities are ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and ${(a - b)^2} = {a^2} - 2ab + {b^2}$. These two identities are further used in another identity which is also important, i.e.
${(a + b)^2} - {(a - b)^2} = 4ab$.