Question

If a=1+√2+√3 and b= 1+√2-√3 prove that a^2-b^2-2a-2b-8=0​

Answer 1

/* There is a mistake in the question. It must be like this */

$Given \: a = 1 + \sqrt{2} + \sqrt{3}$

$\implies a - 1 = \sqrt{2} + \sqrt{3}$

/* On squaring both sides, we get */

$\implies (a-1)^{2} = (\sqrt{2} + \sqrt{3} )^{2}$

$=( \sqrt{2})^{2} + (\sqrt{3})^{2}+2 \times \sqrt{2} \times \sqrt{3}$

$= 2 + 3 + 2\sqrt{6}$

$= 5 +2\sqrt{6} \: --(1)$

$and \: b = 1 + \sqrt{2} - \sqrt{3}$

$\implies b - 1 = \sqrt{2} -\sqrt{3}$

/* On squaring both sides, we get */

$\implies (b-1)^{2} = (\sqrt{2} - \sqrt{3} )^{2}$

$=( \sqrt{2})^{2} + (\sqrt{3})^{2}-2 \times \sqrt{2} \times \sqrt{3}$

$= 2 + 3 - 2\sqrt{6}$

$= 5 - 2\sqrt{6} \: --(2)$

$Now , \red{ (a-1)^{2} + (b-1)^{2} }= 5 +2\sqrt{6} +5 -2\sqrt{6}$

$\implies a^{2} - 2a + 1 + b^{2} - 2b + 1 = 10$

$\implies a^{2} - 2a + b^{2} - 2b + 2-10=0$

$\implies \green { a^{2} + b^{2}-2a - 2b -8 =0}$

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