[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: x = \dfrac{1}{2}\bigg( \sqrt[3]{2009} - \dfrac{1}{ \sqrt[3]{2009} } \bigg) \\ \\ [/tex]
Let assume that
[tex]\bf \: \sqrt[3]{2009} = y \\ \\ [/tex]
So, above equation can be rewritten as
[tex]\sf \: x =\dfrac{1}{2} \bigg(y - \dfrac{1}{y} \bigg) - - - (1) \\ \\ [/tex]
Now, Consider
[tex]\sf \: \sqrt{1 + {x}^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg(y - \dfrac{1}{y} \bigg) }^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \times y \times \frac{1}{y} \bigg) }^{} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \frac{4}{4} + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg(4 + {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \times y \times \frac{1}{y} \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{\dfrac{1}{4} {\bigg(y + \dfrac{1}{y} \bigg) }^{2} } \\ \\ [/tex]
[tex]\sf \: =\dfrac{1}{2} \bigg(y + \dfrac{1}{y} \bigg) \\ \\ [/tex]
So,
[tex]\sf \: \bf\implies \: \sqrt{1 + {x}^{2} } =\dfrac{1}{2} \bigg(y + \dfrac{1}{y} \bigg) - - - (2) \\ \\ [/tex]
[tex]\sf \: {(x + \sqrt{1 + {x}^{2} }) }^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(y - \dfrac{1}{y}\bigg) + \dfrac{1}{2}\bigg(y + \dfrac{1}{y}\bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(y - \dfrac{1}{y} + y + \dfrac{1}{y}\bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(2y \bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {y}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {( \sqrt[3]{2009} )}^{3} \\ \\ [/tex]
[tex]\sf \: = \: 2009 \\ \\ [/tex]
Hence,
[tex]\sf \: \bf\implies \:{(x + \sqrt{1 + {x}^{2} }) }^{3} = 2009 \\ \\ [/tex]
Hence, Option (2) is correct
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Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: x = \dfrac{1}{2}\bigg( \sqrt[3]{2009} - \dfrac{1}{ \sqrt[3]{2009} } \bigg) \\ \\ [/tex]
Let assume that
[tex]\bf \: \sqrt[3]{2009} = y \\ \\ [/tex]
So, above equation can be rewritten as
[tex]\sf \: x =\dfrac{1}{2} \bigg(y - \dfrac{1}{y} \bigg) - - - (1) \\ \\ [/tex]
Now, Consider
[tex]\sf \: \sqrt{1 + {x}^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg(y - \dfrac{1}{y} \bigg) }^{2} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \times y \times \frac{1}{y} \bigg) }^{} } \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{1 + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \frac{4}{4} + \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg(4 + {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{ \dfrac{1}{4} {\bigg( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \times y \times \frac{1}{y} \bigg) }} \\ \\ [/tex]
[tex]\sf \: = \: \sqrt{\dfrac{1}{4} {\bigg(y + \dfrac{1}{y} \bigg) }^{2} } \\ \\ [/tex]
[tex]\sf \: =\dfrac{1}{2} \bigg(y + \dfrac{1}{y} \bigg) \\ \\ [/tex]
So,
[tex]\sf \: \bf\implies \: \sqrt{1 + {x}^{2} } =\dfrac{1}{2} \bigg(y + \dfrac{1}{y} \bigg) - - - (2) \\ \\ [/tex]
Now, Consider
[tex]\sf \: {(x + \sqrt{1 + {x}^{2} }) }^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(y - \dfrac{1}{y}\bigg) + \dfrac{1}{2}\bigg(y + \dfrac{1}{y}\bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(y - \dfrac{1}{y} + y + \dfrac{1}{y}\bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {\bigg[\dfrac{1}{2}\bigg(2y \bigg) \bigg]}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {y}^{3} \\ \\ [/tex]
[tex]\sf \: = \: {( \sqrt[3]{2009} )}^{3} \\ \\ [/tex]
[tex]\sf \: = \: 2009 \\ \\ [/tex]
Hence,
[tex]\sf \: \bf\implies \:{(x + \sqrt{1 + {x}^{2} }) }^{3} = 2009 \\ \\ [/tex]
Hence, Option (2) is correct