Question

If $x = 1 + \sqrt 2$ , then find the value of ${\left( {x - \dfrac{1}{x}} \right)^3}$
A. $6$
B. $7$
C. $8$
D. $10$

Answer 1

To solve this question we need to apply the concept of Binomial expansion and rationalization. Rationalization is the process of eliminating a radical or imaginary number from the denominator of an algebraic fraction. In the first part of the question we will find the value of $\dfrac{1}{x}$ and then subtract it from $x$ . After finding the value we find the cube of the value and we get the result.

Complete step by step answer:
The question ask us to find the value of ${\left( {x - \dfrac{1}{x}} \right)^3}$ when for the “x” is given to us which is a complex number, which is $1 + \sqrt 2$.First we find the value of $\left( {x - \dfrac{1}{x}} \right)$. Now we will find the reciprocal of $x$ number, which means $\dfrac{1}{x}$.

Since we know that the number need to be nationalized so as to remove the complex part of the number from the denominator on rationalizing the above fraction we get $\dfrac{1}{x}$ , which is
$\Rightarrow \dfrac{1}{x} = \dfrac{1}{{1 + \sqrt 2 }}$
On rationalizing we get
$\Rightarrow \dfrac{1}{x} = \dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}$
$\Rightarrow \dfrac{1}{x} = \dfrac{{1 - \sqrt 2 }}{{(1 + \sqrt 2 )(1 - \sqrt 2 )}}$
On analyzing the above expression we see that we can apply the formula $(a + b)(a - b) = {a^2} - {b^2}$ . On applying the same in the denominator of the fraction we get
$\Rightarrow \dfrac{1}{x} = \dfrac{{1 - \sqrt 2 }}{{(1){^2} - {{\left( {\sqrt 2 } \right)}^2}}}$
$\Rightarrow \dfrac{1}{x} = \dfrac{{1 - \sqrt 2 }}{{1 - 2}}$

On calculating further , we get
$\Rightarrow \dfrac{1}{x} = \dfrac{{1 - \sqrt 2 }}{{ - 1}}$
$\Rightarrow \dfrac{1}{x} = - \dfrac{{1 - \sqrt 2 }}{1}$
So, the value of $\dfrac{1}{x}$ becomes $\sqrt 2 - 1$ .
Now, on calculating the expression $x - \dfrac{1}{x}$ , we get
$\Rightarrow x - \dfrac{1}{x} = 1 + \sqrt 2 - \sqrt 2 + 1$
The term $\sqrt 2$ get cancelled so the value becomes
$\Rightarrow x - \dfrac{1}{x} = 1 + 1$
$\Rightarrow x - \dfrac{1}{x} = 2$
Now we take cube both sides of the above equation
$\Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = {2^3}$
$\therefore {\left( {x - \dfrac{1}{x}} \right)^3} = 8$

Hence, the option C is correct.

Note: Remember that the imaginary part in the denominator needs to be changed as the denominator cannot be a complex number. It needs to be changed to a real one so that we will be multiplying both numerators and denominators by the conjugate of the denominator. Conjugate of $a + \sqrt b$ is called $a - \sqrt b$ , the conjugate of a complex number refers to the same real and imaginary number where the sign of the imaginary number is opposite.