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Find the value of \[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}\]

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Last updated date: 15th Jun 2024
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Answer
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Hint: Here, use the basic formula of the difference of two squares and the property of the splitting of the powers using the multiplicative identity. Also, use the fundamental of the squares and square roots which cancels each other.

Complete step-by-step answer:
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}\] ………….. (a)
Simplify using the formula the difference of two squares –
${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$
\[\begin{align}
\Rightarrow & {{\left( {{61}^{2}}-{{11}^{2}} \right)}^{{}}}=(61-11)(61+11) \\
\Rightarrow & \left( {{61}^{2}}-{{11}^{2}} \right)=(50)(72) \\
\Rightarrow & \left( {{61}^{2}}-{{11}^{2}} \right)=3600\,\text{ }..........\text{(b)} \\
\end{align}\](Simplify)
Place the value of equation (b) in the equation (a) –
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}={{(3600)}^{\dfrac{3}{2}}}\] …………..(c)
Simplify, using the basic mathematical operations –
Split the power –
$\Rightarrow$ $\dfrac{3}{2}=\left( \dfrac{1}{2} \right)\times 3$
Place the above value in the equation (c) –
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}={{\left[ {{(3600)}^{\dfrac{1}{2}}} \right]}^{3}}\]
Simplify the left hand side of the equation –
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}={{\left[ {{({{60}^{2}})}^{\dfrac{1}{2}}} \right]}^{3}}\]
By the property – the squares and square-root cancels each other –
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}={{\left[ (60) \right]}^{3}}\]
Now, simplify the left hand side of the equation and do the cubes of six and ten -
\[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}=216000\]
Therefore, the required solution is \[{{\left( {{61}^{2}}-{{11}^{2}} \right)}^{\dfrac{3}{2}}}=216000\]

Note: The square root of the number “n” is the number when multiplied by the number itself and equals to “n”. For example, the square root of $\sqrt{9}=\sqrt{{{3}^{2}}}=3$. The squares and the square roots are opposite to each other and so cancel each other. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $\text{16 = 4 }\times \text{ 4, 16=}{{\text{4}}^{2}}$, generally it is denoted by n to the power two i.e. ${{n}^{2}}$. The perfect square is the number which can be expressed as the product of the two equal integers. For example: $9$, it can be expressed as the product of equal integers. $9=3\times 3$. Cube is the number, we get when the number is multiplied three times. For example - ${{n}^{3}}=n\times n\times n$
${{3}^{3}}=3\times 3\times 3,\ \text{implies }{{3}^{3}}=27$