[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: a = \frac{1}{7 - 4 \sqrt{3} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: a = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ 49 - 16 \times 3} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ 49 - 48} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{1} \\ \\ [/tex]
[tex]\bf\implies \:a = 7 + 4 \sqrt{3} \\ \\ [/tex]
Now, Consider
[tex]\sf \: b = \frac{1}{7 + 4 \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: b = \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{(7)^{2} - (4 \sqrt{3})^{2} } \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{49 - 16 \times 3} \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{49 - 48} \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{1} \\ \\ [/tex]
[tex]\bf\implies \:b = 7 - 4 \sqrt{3} \\ \\ \\ [/tex]
[tex]\sf \: a(a^{2} + 2) + b(b^{2} + 2) \\ \\ [/tex]
[tex]\sf \: = \: {a}^{3} + 2a + {b}^{3} + 2b \\ \\ [/tex]
[tex]\sf \: = \: {a}^{3}+ {b}^{3} + 2a + 2b \\ \\ [/tex]
[tex]\sf \: = \: {(a + b)}^{3} - 3ab(a + b)+ 2(a + b) \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )}^{3} - 3(7 + 4 \sqrt{3})(7 - 4 \sqrt{3}) (7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )+ 2(7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} ) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - 3(49 - 48)(14 )+ 2(14) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - 3(14 )+ 2(14) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - (14) \\ \\ [/tex]
[tex]\sf \: = \: 14[{(14 )}^{2} - 1] \\ \\ [/tex]
[tex]\sf \: = \: 14[{(14 )}^{2} - {1}^{2} ] \\ \\ [/tex]
[tex]\sf \: = \: 14(14 + 1)(14 - 1) \\ \\ [/tex]
[tex]\sf \: = \: 14(15)(13) \\ \\ [/tex]
[tex]\sf \: = \: (2.7)(3.5)(13) \\ \\ [/tex]
[tex]\sf \: = \: 2.3.5.7.13 \\ \\ [/tex]
Hence,
[tex]\implies\boxed{ \bf{ \:a(a^{2} + 2) + b(b^{2} + 2) = 2.3.5.7.13 \: }}\\ \\ [/tex]
So, option (1) is correct.
[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
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: a = \frac{1}{7 - 4 \sqrt{3} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: a = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ 49 - 16 \times 3} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{ 49 - 48} \\ \\ [/tex]
[tex]\sf \: a = \frac{7 + 4 \sqrt{3} }{1} \\ \\ [/tex]
[tex]\bf\implies \:a = 7 + 4 \sqrt{3} \\ \\ [/tex]
Now, Consider
[tex]\sf \: b = \frac{1}{7 + 4 \sqrt{3} } \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: b = \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{(7)^{2} - (4 \sqrt{3})^{2} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{49 - 16 \times 3} \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{49 - 48} \\ \\ [/tex]
[tex]\sf \: b = \frac{7 - 4 \sqrt{3} }{1} \\ \\ [/tex]
[tex]\bf\implies \:b = 7 - 4 \sqrt{3} \\ \\ \\ [/tex]
Now, Consider
[tex]\sf \: a(a^{2} + 2) + b(b^{2} + 2) \\ \\ [/tex]
[tex]\sf \: = \: {a}^{3} + 2a + {b}^{3} + 2b \\ \\ [/tex]
[tex]\sf \: = \: {a}^{3}+ {b}^{3} + 2a + 2b \\ \\ [/tex]
[tex]\sf \: = \: {(a + b)}^{3} - 3ab(a + b)+ 2(a + b) \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }} \\ \\ [/tex]
[tex]\sf \: = \: {(7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )}^{3} - 3(7 + 4 \sqrt{3})(7 - 4 \sqrt{3}) (7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} )+ 2(7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} ) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - 3(49 - 48)(14 )+ 2(14) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - 3(14 )+ 2(14) \\ \\ [/tex]
[tex]\sf \: = \: {(14 )}^{3} - (14) \\ \\ [/tex]
[tex]\sf \: = \: 14[{(14 )}^{2} - 1] \\ \\ [/tex]
[tex]\sf \: = \: 14[{(14 )}^{2} - {1}^{2} ] \\ \\ [/tex]
[tex]\sf \: = \: 14(14 + 1)(14 - 1) \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: = \: 14(15)(13) \\ \\ [/tex]
[tex]\sf \: = \: (2.7)(3.5)(13) \\ \\ [/tex]
[tex]\sf \: = \: 2.3.5.7.13 \\ \\ [/tex]
Hence,
[tex]\implies\boxed{ \bf{ \:a(a^{2} + 2) + b(b^{2} + 2) = 2.3.5.7.13 \: }}\\ \\ [/tex]
So, option (1) is correct.
[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]