Question

If $x - \dfrac{1}{x} = 3$, evaluate ${x^3} - \dfrac{1}{{{x^3}}}.$

Answer 1

Hint: In this given question, first cube on both the sides, then simplify it. We can use the formula of cubic expansion to solve it.

Complete Step by Step Solution:
$x - \dfrac{1}{x} = 3$
Take cube to both the sides of the equation,
$ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = {\left( 3 \right)^3}$
$ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = 27$ . . . (1)
By using the formula of cubic expansion, we ca write
${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2}$
$ \Rightarrow {(a - b)^3} = {a^3} - {b^3} - 3ab(a - b)$
In the above expansion, put
$a = x$ and $b = \dfrac{1}{x}$
$ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = {x^3} - \dfrac{1}{{{x^3}}} - 3 \times x \times \dfrac{1}{x}\left( {x - \dfrac{1}{x}} \right)$ from equation (1)
$ \Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3 \times x \times \dfrac{1}{x}\left( {x - \dfrac{1}{x}} \right) = 27$
Simplifying the above equation, we get
${x^3} - \dfrac{1}{{{x^3}}} - 3 \times 3 = 27$ $\left( {\because x - \dfrac{1}{x} = 3} \right)$
$ \Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 9 = 27$
$ \Rightarrow {x^3} - \dfrac{1}{{{x^3}}} = 27 + 9$
$ \Rightarrow {x^3} - \dfrac{1}{{{x^3}}} = 36$

Note: You can observe that the formula of cubic expansion in standard form was not used in the question. We have manipulated the formula and then applied it. You need to know all the possible manipulations of the formulae to save your time and energy. Knowing just the standard form of any formula is not enough in today’s competition.