If $x+\dfrac{1}{x}=\sqrt{3}$ find the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$
Answer
584.4k+ views
Hint: To obtain the value of given algebraic expression we will use the algebraic identity. Firstly, we have to find the value of an algebraic expression which has a variable with power 3 so we will use the ${{\left( a+b \right)}^{3}}$ formula. Then by using the value given to us we will find the value of the algebraic expression and get the desired answer.
Complete step by step answer:
It is given to us that:
$x+\dfrac{1}{x}=\sqrt{3}$…..$\left( 1 \right)$
We have to find the value of algebraic expression:
${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$…..$\left( 2 \right)$
So as we can see above value has ${{x}^{3}}$ in it so we will use below identity in expression (1):
${{\left( a+b \right)}^{3}}={{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}$
So find the cube of equation (1) as below by using the above formula:
$\begin{align}
& {{\left( x+\dfrac{1}{x} \right)}^{3}}={{\left( \sqrt{3} \right)}^{3}} \\
& \Rightarrow {{x}^{3}}+3\times x\times \dfrac{1}{x}\left( x+\dfrac{1}{x} \right)+{{\left( \dfrac{1}{x} \right)}^{3}}=3\sqrt{3} \\
& \Rightarrow {{x}^{3}}+3\left( x+\dfrac{1}{x} \right)+\dfrac{1}{{{x}^{3}}}=3\sqrt{3} \\
\end{align}$
Now from we will replace the value in above expression from equation (1) as follows:
$\begin{align}
& {{x}^{3}}+3\left( \sqrt{3} \right)+\dfrac{1}{{{x}^{3}}}=3\sqrt{3} \\
& \Rightarrow {{x}^{3}}+\dfrac{1}{{{x}^{3}}}=3\sqrt{3}-3\sqrt{3} \\
\end{align}$
$\therefore {{x}^{3}}+\dfrac{1}{{{x}^{3}}}=0$
Hence the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$ is 0.
Note: Algebraic expressions are the expressions that are made up of variables and constant along with the algebraic operation such as addition, subtraction, multiplication and division. On the other hand algebraic equations are the equations in which two expressions are set equal to each other. The difference between the two is that algebraic expressions don’t have equal sign and algebraic equations have equal sign. The algebraic equations that are valid for any value of the variable in it are known as algebraic identities. Algebraic identity is a standard form that can be used for any value of the variables. They are used to solve the different polynomials and the algebraic equations.
Complete step by step answer:
It is given to us that:
$x+\dfrac{1}{x}=\sqrt{3}$…..$\left( 1 \right)$
We have to find the value of algebraic expression:
${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$…..$\left( 2 \right)$
So as we can see above value has ${{x}^{3}}$ in it so we will use below identity in expression (1):
${{\left( a+b \right)}^{3}}={{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}$
So find the cube of equation (1) as below by using the above formula:
$\begin{align}
& {{\left( x+\dfrac{1}{x} \right)}^{3}}={{\left( \sqrt{3} \right)}^{3}} \\
& \Rightarrow {{x}^{3}}+3\times x\times \dfrac{1}{x}\left( x+\dfrac{1}{x} \right)+{{\left( \dfrac{1}{x} \right)}^{3}}=3\sqrt{3} \\
& \Rightarrow {{x}^{3}}+3\left( x+\dfrac{1}{x} \right)+\dfrac{1}{{{x}^{3}}}=3\sqrt{3} \\
\end{align}$
Now from we will replace the value in above expression from equation (1) as follows:
$\begin{align}
& {{x}^{3}}+3\left( \sqrt{3} \right)+\dfrac{1}{{{x}^{3}}}=3\sqrt{3} \\
& \Rightarrow {{x}^{3}}+\dfrac{1}{{{x}^{3}}}=3\sqrt{3}-3\sqrt{3} \\
\end{align}$
$\therefore {{x}^{3}}+\dfrac{1}{{{x}^{3}}}=0$
Hence the value of ${{x}^{3}}+\dfrac{1}{{{x}^{3}}}$ is 0.
Note: Algebraic expressions are the expressions that are made up of variables and constant along with the algebraic operation such as addition, subtraction, multiplication and division. On the other hand algebraic equations are the equations in which two expressions are set equal to each other. The difference between the two is that algebraic expressions don’t have equal sign and algebraic equations have equal sign. The algebraic equations that are valid for any value of the variable in it are known as algebraic identities. Algebraic identity is a standard form that can be used for any value of the variables. They are used to solve the different polynomials and the algebraic equations.
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