where a and b are natural number find (a + b).

1.	where a and b are natural number find (a + b).
Answer» $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$ $\boxed{ \sf \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$ $\large\underline{\sf{Solution-}}$ CONSIDER, $\sf \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{13 + \sqrt{48} } } } \bigg)$ $\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{12 + 1 + \sqrt{2 \times 2 \times 4 \times 3} } } } \bigg)$ $\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{ {( \sqrt{12)} }^{2} + {(1)}^{2} + 2 \sqrt{12} } } } \bigg)$ $\sf \: = 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{ {\bigg( \sqrt{12} + 1 \bigg) }^{2} } } } \bigg)$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy }$ $\sf \: = 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{12} - 1 } } \bigg)$ $\sf \: = 2\bigg( \sqrt{3 + \sqrt{4 - \sqrt{12} } } \bigg)$ $\sf \: = 2\bigg( \sqrt{3 + \sqrt{3 + 1 - \sqrt{2 \times 2 \times 3}} } \bigg)$ $\sf \: = 2\bigg( \sqrt{3 + \sqrt{ {( \sqrt{3)} }^{2} + {(1)}^{2} - 2 \sqrt{3}} } \bigg)$ $\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{ {( \sqrt{3} - 1) }^{2} } } \bigg)$ $\sf \: = \: 2 \sqrt{\bigg(3 + \sqrt{3} - 1 \bigg) }$ $\sf \: = \: 2 \sqrt{\bigg(2 + \sqrt{3} \bigg) }$ $\sf \: = \: 2\sqrt{ \bigg( \dfrac{3}{2} + \dfrac{1}{2} + \sqrt{3} \times \dfrac{2}{ \sqrt{2} \times \sqrt{2} } \bigg) }$ $\sf \: = \: 2 \sqrt{\bigg( {\bigg(\dfrac{ \sqrt{3} }{ \sqrt{2} } \bigg) }^{2} + {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2} + 2 \times \dfrac{ \sqrt{3} }{ \sqrt{2} } \times \dfrac{1}{ \sqrt{2} } \bigg) }$ $\sf \: = \: 2 \sqrt{ {\bigg(\dfrac{ \sqrt{3} }{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } \bigg) }^{2} }$ $\sf \: = \: 2\bigg( \dfrac{ \sqrt{3} }{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } \bigg)$ $\sf \: = \: \sqrt{2} \times \sqrt{2} \times \bigg(\dfrac{ \sqrt{3} }{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } \bigg)$ $\sf \: = \: \sqrt{6} + \sqrt{2}$ So, $\rm :\implies\:\sf \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{13 + \sqrt{48} } } } \bigg) = \sqrt{6} + \sqrt{2}$ As it is GIVEN that, $\sf \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{13 + \sqrt{48} } } } \bigg) = \sqrt{a} + \sqrt{b}$ So, On COMPARING we GET, $\rm :\implies\: \sqrt{a} + \sqrt{b} = \sqrt{6} + \sqrt{2}$ $\bf\implies \:a = 6 \: and \: b = 2 \: \: or \: a = 2 \: and \: b = 6$ So, $\bf\implies \:a + b = 2 + 6 = 8$ $\overbrace{ \underline { \boxed { \rm \therefore \: The \: value \: of \: a \: + \: b \: = \: 8 }}}$

Answer»

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$\boxed{ \sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\boxed{ \sf \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$

$\large\underline{\sf{Solution-}}$

CONSIDER,

$\sf \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{13 + \sqrt{48} } } } \bigg)$

$\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{12 + 1 + \sqrt{2 \times 2 \times 4 \times 3} } } } \bigg)$

$\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{ {( \sqrt{12)} }^{2} + {(1)}^{2} + 2 \sqrt{12} } } } \bigg)$

$\sf \: = 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{ {\bigg( \sqrt{12} + 1 \bigg) }^{2} } } } \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy }$

$\sf \: = 2\bigg( \sqrt{3 + \sqrt{5 - \sqrt{12} - 1 } } \bigg)$

$\sf \: = 2\bigg( \sqrt{3 + \sqrt{4 - \sqrt{12} } } \bigg)$

$\sf \: = 2\bigg( \sqrt{3 + \sqrt{3 + 1 - \sqrt{2 \times 2 \times 3}} } \bigg)$

$\sf \: = 2\bigg( \sqrt{3 + \sqrt{ {( \sqrt{3)} }^{2} + {(1)}^{2} - 2 \sqrt{3}} } \bigg)$

$\sf \: = \: 2\bigg( \sqrt{3 + \sqrt{ {( \sqrt{3} - 1) }^{2} } } \bigg)$

$\sf \: = \: 2 \sqrt{\bigg(3 + \sqrt{3} - 1 \bigg) }$

$\sf \: = \: 2 \sqrt{\bigg(2 + \sqrt{3} \bigg) }$

$\sf \: = \: 2\sqrt{ \bigg( \dfrac{3}{2} + \dfrac{1}{2} + \sqrt{3} \times \dfrac{2}{ \sqrt{2} \times \sqrt{2} } \bigg) }$

$\sf \: = \: 2 \sqrt{\bigg( {\bigg(\dfrac{ \sqrt{3} }{ \sqrt{2} } \bigg) }^{2} + {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2} + 2 \times \dfrac{ \sqrt{3} }{ \sqrt{2} } \times \dfrac{1}{ \sqrt{2} } \bigg) }$