Question

If \[x+\dfrac{1}{x}=5\] , then find \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\].

Answer 1

Hint: As the data we are given and what we have to calculate is of the same form, and on seeing carefully we conclude that there is some relation between them and we need to just modify the question. So, the required value \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\] can be written in terms of \[{{(x+\dfrac{1}{x})}^{3}}\] since there is exponential power \[3\]. The required value is written as the \[cube\] of \[x+\dfrac{1}{x}\] and minus something and on writing down we see that the minus term has some relation between the given data and that can be written as thrice of \[x+\dfrac{1}{x}\].

Complete step by step solution:
Given that, \[x+\dfrac{1}{x}=5\]
\[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}=?\]
As the given data and required data is of the same form means there must be some relation between them as we don’t have any other data to solve.
On seeing the exponential power as \[3\] ,
Also, we know that \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3({{a}^{2}}b+a{{b}^{2}})\]
\[\Rightarrow {{(x+\dfrac{1}{x})}^{3}}={{x}^{3}}+\dfrac{1}{{{x}^{3}}}+3(x+\dfrac{1}{x})\]
\[\Rightarrow {{x}^{3}}+\dfrac{1}{{{x}^{3}}}={{(x+\dfrac{1}{x})}^{3}}-3(x+\dfrac{1}{x})\]
Since, \[x+\dfrac{1}{x}=5\]
On substituting this value in above equation
\[\Rightarrow {{x}^{3}}+\dfrac{1}{{{x}^{3}}}={{(5)}^{3}}-3(5)\]
\[\Rightarrow 125-15\]
\[\Rightarrow 110\]

Thus, we have calculated the value of \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}\]
Hence \[{{x}^{3}}+\dfrac{1}{{{x}^{3}}}=110\]

Note:
In this type of questions firstly note down the given data and just think about the relation between what we are given and what we need to calculate and also must remember the algebraic formulae/results. In this type of questions just remember that we need to raise the same exponential power in the given data as in the required value.