If 19+837−43=a+b3, then 'a'=

19+837−43=a+b319+2487−212=16+3+216×34+3−24×3⇒16+34−3=4+32−3×2+32+3=(4+3)(2+3)⇒8+3+23+63=a+b3⇒11+63=a+b3a=11,b=6

If 19+837−43=a+b3, then 'a'=

19+837−43=a+b319+2487−212=16+3+216×34+3−24×3⇒16+34−3=4+32−3×2+32+3=(4+3)(2+3)⇒8+3+23+63=a+b3⇒11+63=a+b3a=11,b=6

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If $\sqrt{\frac{19 + 8 \sqrt{3}}{7 - 4 \sqrt{3}}} = a + b \sqrt{3}$ , then $' a' =$

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$\sqrt{\frac{19 + 8 \sqrt{3}}{7 - 4 \sqrt{3}}} = a + b \sqrt{3}$

$\frac{\sqrt{19 + 2 \sqrt{48}}}{\sqrt{7 - 2 \sqrt{12}}} = \frac{\sqrt{16 + 3 + 2 \sqrt{16 \times 3}}}{\sqrt{4 + 3 - 2 \sqrt{4 \times 3}}}$

$\Rightarrow \frac{\sqrt{16} + \sqrt{3}}{\sqrt{4} - \sqrt{3}} = \frac{4 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$

$= (4 + \sqrt{3}) (2 + \sqrt{3})$

$\Rightarrow 8 + 3 + 2 \sqrt{3} + 6 \sqrt{3} = a + b \sqrt{3}$

$\Rightarrow 11 + 6 \sqrt{3} = a + b \sqrt{3}$

$a = 11, b = 6$

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If $\sqrt{\frac{19 + 8 \sqrt{3}}{7 - 4 \sqrt{3}}} = a + b \sqrt{3}$ , then $' a' =$