Question

How do you simplify $\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}}$ ?

Answer 1

Hint:To solve this equation, we let the correct expression of the equation to be $\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}}$In this question, we are given a fraction containing numerical values and some unknown variable quantity “x” that is raised to the power “2n” in the numerator and “n” in the denominator. For simplifying this fraction, we will use the identities of the algorithm, then we will look for the factors that are present in both the numerator and the denominator, these factors are called the common factors, then we cancel out the common factors until there are no common factors present between the numerator and the denominator.

Complete step by step answer:
We have to simplify $\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}}$
It can be rewritten as –
$\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{{{({x^n})}^2} - {{(1)}^2}}}{{{x^n} - 3}}$
We know that
$
{a^2} - {b^2} = (a - b)(a + b) \\
\Rightarrow {({x^n})^2} - {(1)^2} = ({x^n} - 1)({x^n} + 1) \\
$
Using this value in the given expression, we get –
$\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 1)({x^n} + 1)}}{{{x^n} - 3}}$
Now, -1 can be written as $ - 3 + 2$ , so we get –
$
\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3 + 2)({x^n} + 1)}}{{{x^n} - 3}} \\
\Rightarrow \dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3)({x^n} + 1) + 2({x^n} + 1)}}{{{x^n} -
3}} \\
$
Now, 1 can be written as $4 - 3$ , so we get –
$
\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3)({x^n} + 1) + 2({x^n} + 4 - 3)}}{{{x^n} - 3}} \\
\Rightarrow \dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3)({x^n} + 1) + 2({x^n} - 3) + 8}}{{{x^n}
- 3}} \\
\Rightarrow \dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3)({x^n} + 1) + 2({x^n} - 3)}}{{{x^n} -
3}} + \dfrac{8}{{{x^n} - 3}} \\
$
Now, we take $({x^n} - 3)$ common in the numerator and get –
$\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = \dfrac{{({x^n} - 3)[{x^n} + 1 + 2]}}{{{x^n} - 3}} + \dfrac{8}{{{x^n} -
3}}$
${x^n} - 3$ is present in both the numerator and the denominator, so we cancel it out –
$ \Rightarrow \dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}} = {x^n} + 3 + \dfrac{8}{{{x^n} - 3}}$
Hence the simplified form of $\dfrac{{{x^{2n}} - 1}}{{{x^n} - 3}}$ is ${x^n} + 3 + \dfrac{8}{{{x^n} - 3}}$ .

Note: In this solution, we have used a law of exponential which states that when a number that is already raised to some power is again raised to some power then the base remains the same and the powers are multiplied with each other and vice versa, that is, ${({x^n})^m} = {x^{n \times m}}$ or ${x^{n
\times m}} = {({x^n})^m}$ . We have also used one algorithmic identity and the distributive property several identities several times in this solution.