[tex]\text{$(t=\sqrt[3]{2009})$}[/tex]
[tex]\;[/tex]
[tex]\text{$x=\dfrac{1}{2}\bigg(t-\dfrac{1}{t}\bigg)$}[/tex]
To obtain [tex]\text{$x^{2}$}[/tex], we square both sides.
[tex]\text{$x^{2}=\dfrac{1}{4}\bigg(t-\dfrac{1}{t}\bigg)^{2}$}[/tex]
[tex]\text{$x^{2}=\dfrac{1}{4}\bigg(t^{2}-2+\dfrac{1}{t^{2}}\bigg)$}[/tex]
To obtain [tex]\text{$\sqrt{1+x^{2}}$}[/tex], we add 1 on both sides.
[tex]\text{$1+x^{2}=\dfrac{1}{4}\bigg(t^{2}+2+\dfrac{1}{t^{2}}\bigg)$}[/tex]
We notice the right-hand side is a perfect square.
[tex]\text{$1+x^{2}=\bigg(\dfrac{1}{2}\bigg)^{2}\bigg(t+\dfrac{1}{t}\bigg)^{2}$}[/tex]
[tex]\text{$\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg|t+\dfrac{1}{t}\bigg|$}[/tex]
[tex]\text{$\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg(t+\dfrac{1}{t}\bigg)$}[/tex]
Now, add the rest to obtain [tex]\text{$(x+\sqrt{1+x^{2}})^{3}$}[/tex].
[tex]\text{$x+\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg(t+t+\dfrac{1}{t}-\dfrac{1}{t}\bigg)$}[/tex]
[tex]\text{$x+\sqrt{1+x^{2}}=t$}[/tex]
[tex]\text{$(x+\sqrt{1+x^{2}})^{3}=t^{3}$}[/tex]
[tex]\text{$\therefore (x+\sqrt{1+x^{2}})^{3}=\boxed{2009}$}[/tex]
Option 2 is the correct choice.
Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:(2) \: \: 2009 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Given that
[tex]\sf \: x= \dfrac{1}{2} \left( \sqrt[3]{2009} - \dfrac{1}{ \sqrt[3]{2009} } \right) \\ \\ [/tex]
Let assume that
[tex]\qquad\sf \: \sqrt[3]{2009} = y \\ \\ [/tex]
So, above expression can be rewritten as
[tex]\sf\implies \bf \: x = \dfrac{1}{2} \left(y - \dfrac{1}{y} \right) - - - (1)\\ \\ [/tex]
Now,
[tex]\sf \: {x}^{2} = \dfrac{1}{4} \left(y - \dfrac{1}{y} \right)^{2} \\ \\ [/tex]
[tex]\sf \: {x}^{2} = \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
Now, Consider
[tex]\sf \: 1 + {x}^{2} = 1 + \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left(4 + {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left(y + \dfrac{1}{y} \right)^{2} \\ \\ [/tex]
So,
[tex]\sf\implies \bf \: \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(y + \dfrac{1}{y} \right) - - - (2) \\ \\ [/tex]
Now, Adding equation (1) and (2), we get
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(y - \dfrac{1}{y} + y + \dfrac{1}{y} \right) \\ \\ [/tex]
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(2y\right) \\ \\ [/tex]
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = y \\ \\ [/tex]
On cubing both sides, we get
[tex]\sf \: (x + \sqrt{1 + {x}^{2} })^{3} = {y}^{3} \\ \\ [/tex]
So, on substituting the value of y, we get
[tex]\sf \: (x + \sqrt{1 + {x}^{2} })^{3} = {( \sqrt[3]{2009} )}^{3} \\ \\ [/tex]
[tex]\sf\implies \bf \: (x + \sqrt{1 + {x}^{2} })^{3} =2009 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Answers & Comments
Verified answer
Solution
[tex]\text{$(t=\sqrt[3]{2009})$}[/tex]
[tex]\;[/tex]
[tex]\text{$x=\dfrac{1}{2}\bigg(t-\dfrac{1}{t}\bigg)$}[/tex]
To obtain [tex]\text{$x^{2}$}[/tex], we square both sides.
[tex]\text{$x^{2}=\dfrac{1}{4}\bigg(t-\dfrac{1}{t}\bigg)^{2}$}[/tex]
[tex]\text{$x^{2}=\dfrac{1}{4}\bigg(t^{2}-2+\dfrac{1}{t^{2}}\bigg)$}[/tex]
To obtain [tex]\text{$\sqrt{1+x^{2}}$}[/tex], we add 1 on both sides.
[tex]\text{$1+x^{2}=\dfrac{1}{4}\bigg(t^{2}+2+\dfrac{1}{t^{2}}\bigg)$}[/tex]
We notice the right-hand side is a perfect square.
[tex]\text{$1+x^{2}=\bigg(\dfrac{1}{2}\bigg)^{2}\bigg(t+\dfrac{1}{t}\bigg)^{2}$}[/tex]
[tex]\text{$\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg|t+\dfrac{1}{t}\bigg|$}[/tex]
[tex]\text{$\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg(t+\dfrac{1}{t}\bigg)$}[/tex]
Now, add the rest to obtain [tex]\text{$(x+\sqrt{1+x^{2}})^{3}$}[/tex].
[tex]\text{$x+\sqrt{1+x^{2}}=\dfrac{1}{2}\bigg(t+t+\dfrac{1}{t}-\dfrac{1}{t}\bigg)$}[/tex]
[tex]\text{$x+\sqrt{1+x^{2}}=t$}[/tex]
[tex]\text{$(x+\sqrt{1+x^{2}})^{3}=t^{3}$}[/tex]
[tex]\text{$\therefore (x+\sqrt{1+x^{2}})^{3}=\boxed{2009}$}[/tex]
[tex]\;[/tex]
Option 2 is the correct choice.
Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:(2) \: \: 2009 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Given that
[tex]\sf \: x= \dfrac{1}{2} \left( \sqrt[3]{2009} - \dfrac{1}{ \sqrt[3]{2009} } \right) \\ \\ [/tex]
Let assume that
[tex]\qquad\sf \: \sqrt[3]{2009} = y \\ \\ [/tex]
So, above expression can be rewritten as
[tex]\sf\implies \bf \: x = \dfrac{1}{2} \left(y - \dfrac{1}{y} \right) - - - (1)\\ \\ [/tex]
Now,
[tex]\sf \: {x}^{2} = \dfrac{1}{4} \left(y - \dfrac{1}{y} \right)^{2} \\ \\ [/tex]
[tex]\sf \: {x}^{2} = \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
Now, Consider
[tex]\sf \: 1 + {x}^{2} = 1 + \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left(4 + {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left( {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \right) \\ \\ [/tex]
[tex]\sf \: 1 + {x}^{2} = \dfrac{1}{4} \left(y + \dfrac{1}{y} \right)^{2} \\ \\ [/tex]
So,
[tex]\sf\implies \bf \: \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(y + \dfrac{1}{y} \right) - - - (2) \\ \\ [/tex]
Now, Adding equation (1) and (2), we get
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(y - \dfrac{1}{y} + y + \dfrac{1}{y} \right) \\ \\ [/tex]
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = \dfrac{1}{2} \left(2y\right) \\ \\ [/tex]
[tex]\sf \: x + \sqrt{1 + {x}^{2} } = y \\ \\ [/tex]
On cubing both sides, we get
[tex]\sf \: (x + \sqrt{1 + {x}^{2} })^{3} = {y}^{3} \\ \\ [/tex]
So, on substituting the value of y, we get
[tex]\sf \: (x + \sqrt{1 + {x}^{2} })^{3} = {( \sqrt[3]{2009} )}^{3} \\ \\ [/tex]
[tex]\sf\implies \bf \: (x + \sqrt{1 + {x}^{2} })^{3} =2009 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]