Question

If $x+\dfrac{1}{x}=5$ then find the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$.

Answer 1

Hint:In this first we have to take whole cubes of the given expression, then apply appropriate algebraic identity to solve the unknown expressions.

Complete step-by-step answer:
Given expression is
              $x+\dfrac{1}{x}=5$
On cubing both sides, we get
           ${{\left( x+\dfrac{1}{x} \right)}^{3}}={{5}^{3}}$
 It is clear from the above equation the left side term is in the form of algebraic identity ${{\left( a+b \right)}^{3}}$.
Now, we have to compare the given term with the suitable identity, then we get
  $a=x\,\,and\,\,b=\dfrac{1}{x}$
As we know that the expanded form of
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
On putting value of a and b in the expanded form of identity, we get
${{x}^{3}}+{{\left( \dfrac{1}{x} \right)}^{3}}+3x\times \dfrac{1}{x}\left( x+\dfrac{1}{x} \right)={{5}^{3}}$
Now on putting the value of $x+\dfrac{1}{x}=5$ and to cancel out x to x , $x\times \dfrac{1}{x}=1$and ${{5}^{3}}=125$ we get
${{x}^{3}}+{{\left( \dfrac{1}{x} \right)}^{3}}+3\times 5=125$
${{x}^{3}}+{{\left( \dfrac{1}{x} \right)}^{3}}+15=125$
On transferring 15 right side, we get
${{x}^{3}}+{{\left( \dfrac{1}{x} \right)}^{3}}=110$
Hence the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}=110$

Additional Information:Algebraic identity- Some equations of algebra which are true for all values of variables are called algebraic identity. Polynomials can be factored using algebraic identity.
There are so many algebraic identities in mathematics.
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

Note:In such types of problems , we need to take care of algebraic identities like which identity is suitable for the given question and what is the expanded form of identities. These questions are also solved by factorization ,if we know how to factorize the given polynomial then it becomes more easy.Students should remember the basic algebraic identities for solving these types of questions.