Question

If we have an algebraic expression $x-\dfrac{1}{x}=5$ , Then find the value of ${{x}^{3}}-\dfrac{1}{{{x}^{3}}}$.
(a) 167
(b) 169
(c) 140
(d) 160

Answer 1

Hint: Apply general algebraic identity, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. Substitute the value of a as x and then substitute value of b as \[\dfrac{1}{x}\]. From this you can simplify the identity by general algebraic methods like squaring on both sides. After squaring just cancel x on numerator and denominator, from which you can easily get the required expression.

Complete step-by-step solution -
By general knowledge of algebra, we can say that:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
Now if we assume
$a=x,b=\dfrac{1}{x}$
By substituting these we get
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=\left( x-\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{x}.x+\dfrac{1}{{{x}^{2}}} \right)$
By simplifying we get
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=\left( x-\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+1 \right)..........(ii)$
Now, we get the relation between the required expression and the given expression. What we need is the remaining expression.
Given expression in the question:
$x-\dfrac{1}{x}=5$
By squaring on both sides, we get:
${{\left( x-\dfrac{1}{x} \right)}^{2}}=25$
By writing it in different form, we get:
$\left( x-\dfrac{1}{x} \right)\left( x-\dfrac{1}{x} \right)=25$
By using distributive law:
$a\left( b+c \right)=a.b+a.c$
By applying above law, we get:
$x.x+\dfrac{1}{x}.\dfrac{1}{x}-x.\dfrac{1}{x}-x.\dfrac{1}{x}=25$
By simplifying, we get:
${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-1-1=25$
By adding 2 on both sides, we get:
${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2+2=27$
By simplifying, we get:
${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27.........(iii)$
Given expression in the question:
$x -\dfrac{1}{x}=5$
By solving we get equation (ii) as:
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=\left( x-\dfrac{1}{x} \right)\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+1 \right)$
By solving we got equation (iii) as:
${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27$
By adding 1 on both sides, we get:
$\begin{align}
  & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+1=27+1 \\
 & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+1=28............(iv) \\
\end{align}$
By substituting equation (iv), given expression into equation (ii) we get:
${{x}^{3}}-\dfrac{1}{{{x}^{3}}}=5\times 28=130+10=140$
So, the value of the required expression is 140.
Therefore, the required value is 140.
Option (c) is correct

Note: Alternate method is instead of using distributive law at (a-b)(a-b) you can write it as a square of the algebraic expression (a-b). Now use the general algebraic identity
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
By distributive law also you get the same answer. But using this makes us get the required result in a single step.