
Find the value :${{\left( \sqrt{32}-\sqrt{5} \right)}^{\dfrac{1}{3}}}{{\left( \sqrt{32}+\sqrt{5} \right)}^{\dfrac{1}{3}}}$.
Answer
572.1k+ views
Hint:For this question, first we have to apply appropriate exponential law to calcite the expression, after that observe the question carefully, then apply suitable algebraic identity to solve it .
Complete step-by-step answer:
The given expression is
${{\left( \sqrt{32}-\sqrt{5} \right)}^{\dfrac{1}{3}}}{{\left( \sqrt{32}+\sqrt{5} \right)}^{\dfrac{1}{3}}}$
To solve this first we have to find which exponential law is using here ,for this we will be using
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
If we have different bases and same exponent , then we will rearrange the given question as above law, so that we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}.....................\left( i \right)$
Now , if we observe the obtained term is the form of an algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
On comparing the equation (i) with the above identity, we get
$a=\sqrt{32\,,}\,\,b=\sqrt{5}$
On putting these values in the identity we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{[{{\left( \sqrt{32} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}]}^{\dfrac{1}{3}}}$
Now, on applying this law ${{\left( \sqrt{a} \right)}^{2}}={{\left( {{a}^{\dfrac{1}{2}}} \right)}^{2}}=a$ we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{\left( 32-5 \right)}^{\dfrac{1}{3}}}$
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{\left( 27 \right)}^{\dfrac{1}{3}}}\,\,\,......................\left( ii \right)$
We know that $27$ is a cube of $3$. Therefore we can write it as
${{\left( 27 \right)}^{\dfrac{1}{3}}}={{\left( {{3}^{3}} \right)}^{\dfrac{1}{3}}}=3$……………..(iii)
The above term is also part of the exponential equation.
By equation (ii) and (iii), we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}=3$
Hence, the value of ${{\left( \sqrt{32}-\sqrt{5} \right)}^{\dfrac{1}{3}}}{{\left( \sqrt{32}+\sqrt{5} \right)}^{\dfrac{1}{3}}}= 3$
Additional Information:Exponent- A quantity representing the power to which a given number or expression is to be raised. Exponents are also called Powers or indices.
Laws of exponent-To solve the exponential expression we have 6-7 laws that can help us to solve the given expression. Some basic laws are
$\begin{align}
& \left( 1 \right){{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\
& \left( 2 \right){{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} \\
\end{align}$
$\left( 3 \right){{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$
Note:In such types of problems, it's mandatory to solve step by step with the help of exponents law and identity. Sometimes students get confused and apply the simple multiplication from one bracket to another bracket which will further lead to difficulty to solve the question.
Complete step-by-step answer:
The given expression is
${{\left( \sqrt{32}-\sqrt{5} \right)}^{\dfrac{1}{3}}}{{\left( \sqrt{32}+\sqrt{5} \right)}^{\dfrac{1}{3}}}$
To solve this first we have to find which exponential law is using here ,for this we will be using
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
If we have different bases and same exponent , then we will rearrange the given question as above law, so that we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}.....................\left( i \right)$
Now , if we observe the obtained term is the form of an algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
On comparing the equation (i) with the above identity, we get
$a=\sqrt{32\,,}\,\,b=\sqrt{5}$
On putting these values in the identity we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{[{{\left( \sqrt{32} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}]}^{\dfrac{1}{3}}}$
Now, on applying this law ${{\left( \sqrt{a} \right)}^{2}}={{\left( {{a}^{\dfrac{1}{2}}} \right)}^{2}}=a$ we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{\left( 32-5 \right)}^{\dfrac{1}{3}}}$
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}={{\left( 27 \right)}^{\dfrac{1}{3}}}\,\,\,......................\left( ii \right)$
We know that $27$ is a cube of $3$. Therefore we can write it as
${{\left( 27 \right)}^{\dfrac{1}{3}}}={{\left( {{3}^{3}} \right)}^{\dfrac{1}{3}}}=3$……………..(iii)
The above term is also part of the exponential equation.
By equation (ii) and (iii), we get
${{\left\{ \left( \sqrt{32}-\sqrt{5} \right)\left( \sqrt{32}+\sqrt{5} \right) \right\}}^{\dfrac{1}{3}}}=3$
Hence, the value of ${{\left( \sqrt{32}-\sqrt{5} \right)}^{\dfrac{1}{3}}}{{\left( \sqrt{32}+\sqrt{5} \right)}^{\dfrac{1}{3}}}= 3$
Additional Information:Exponent- A quantity representing the power to which a given number or expression is to be raised. Exponents are also called Powers or indices.
Laws of exponent-To solve the exponential expression we have 6-7 laws that can help us to solve the given expression. Some basic laws are
$\begin{align}
& \left( 1 \right){{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\
& \left( 2 \right){{a}^{m}}\div {{a}^{n}}={{a}^{m-n}} \\
\end{align}$
$\left( 3 \right){{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$
Note:In such types of problems, it's mandatory to solve step by step with the help of exponents law and identity. Sometimes students get confused and apply the simple multiplication from one bracket to another bracket which will further lead to difficulty to solve the question.
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