Question

$[( \sqrt{3} - \sqrt{2} +1) ^{2} -2(3- \sqrt{6}) ]* ( \sqrt{12} + \sqrt{8} )$

Answer 1

(3-2*rad6+2+1-2(3-rad6)*(rad12+rad8)=
(2(3-rad6)-2(3-rad6)*(rad12+rad18)=0*(rad12+rad8)=0

Answer 2

Pntru paranteza
$( \sqrt{3} - \sqrt{2}+1 )^{2}$ folosim formula pentru
$( a - b+c )^{2} = a^{2} + b^{2} + c^{2} + 2*a*b + 2*a*c + 2*b*c$ si obtinem:
(3+2+1-$2* \sqrt{6}$ + $2* \sqrt{3}$ - $2* \sqrt{2}$ -6 + $2* \sqrt{6}$)*($2* \sqrt{3}$ + $2* \sqrt{2}$)=

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

=$2*( \sqrt{3}$ - $\sqrt{2})$)*$2* (\sqrt{3}$ + $\sqrt{2})$ =
=4*(3-2)=4*1=4