Question

Hello everyone....



Please solve it
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Answer 1

$\bold{\underline{Hey\: mates !}}$​



$[\sqrt{3}+\sqrt{2}]^{6} - [\sqrt{3}-\sqrt{2}]^{6}$



Now ,



$a = [\sqrt{3}+\sqrt{2}]$


$b = [\sqrt{3}-\sqrt{2}]$



We have ,



$x^{3}-y^{3} = (x-y)(x^{2} + xy + y^{2})$



$\therefore\: a^{6} - b^{6} \\\\\implies (a^{2})^{3} - (a^{2})^{3} \\\\\implies [a^2 - b^2][ a^{2} +a^{2}b^{2} + b^{2}]$



Now put the value of a & b



$\implies[(\sqrt{3}+\sqrt{2})^{2} -(\sqrt{3}+\sqrt{2})^{2}][((\sqrt{3}+\sqrt{2})^{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}(\sqrt{3}-\sqrt{2})^{2}+((\sqrt{3}-\sqrt{2})^{2})^{2}]$



$\implies [ 3 + 2 + 2\sqrt{6} - 3 - 2 + 2\sqrt{6}][(3+2+2\sqrt{6})^{2} +(3+2-2\sqrt{6})^{2} (3+2-2\sqrt{6})^{2} + (3+2-2\sqrt{6})^{2} ]$



$\implies [ 4\sqrt{6} ][ (5+2\sqrt{6})^{2} + (5+2\sqrt{6})(5-2\sqrt{6})+(5-2\sqrt{6})^{2}$



$\implies [ 4\sqrt{6} ][ 25 + 24 + \cancel{20\sqrt{6}} + ( 5^{2} - (2\sqrt{6})^{2} ) + 25 +24 - \cancel{20\sqrt{6}}]$



$\implies [4\sqrt{6} ][ 98 + ( 25 -24)]$



$\implies [4\sqrt{6} ][ 98 + 1 ]$



$\implies [4\sqrt{6} ][ 99]$


$\implies [396\sqrt{6}]$


$\bold{\underline{Answer}}$


$\boxed{[\sqrt{3}+\sqrt{2}]^{6} - [\sqrt{3}-\sqrt{2}]^{6} = 396\sqrt{6}}$



$<marquee = right> Thanks$

Answer 2

Answer:

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