Помогите,пожалуйста! С решением,если можно)

$\sqrt{21-2\sqrt{21+2\sqrt{19-6\sqrt2}}}\cdot (3\sqrt2+1)=17;$

$1)19-6\sqrt2=19-2\cdot 3\sqrt2=18+1-2\cdot 3\sqrt2=(9\cdot 2)-2\cdot 3\sqrt2+1=\\\\=(3\sqrt2)^2-2\cdot 3\sqrt2\cdot 1+1=(3\sqrt2-1)^2\\\\\\2)\sqrt{21+2\sqrt{19-6\sqrt2}}=\sqrt{21+2\sqrt{(3\sqrt2-1)^2}}=\sqrt{21+2(3\sqrt2-1)}=\\\\=\sqrt{21+2\cdot 3\sqrt3-2}=\sqrt{19+2\cdot 3\sqrt2}=\sqrt{(3\sqrt2+1)^2}=3\sqrt2+1\\\\\\3)\sqrt{21-2\sqrt{(3\sqrt2+1)^2}}*(3\sqrt2+1)=\sqrt{21-2(3\sqrt2+1)}*(3\sqrt2+1)=\\\\=\sqrt{21-2\cdot 3\sqrt2-2}*(3\sqrt2+1)=\sqrt{19-2\cdot 3\sqrt2}*(3\sqrt2+1)=\sqrt{(3\sqrt2-1)^2}*(3\sqrt2+1)=\\\\=(3\sqrt2-1)*(3\sqrt2+1)=9*2-1=17;$

0\\\\\sqrt{(3\sqrt2+1)^2}=|3\sqrt2+1|=3\sqrt2+1,t.k.3\sqrt2+1>0" alt="P.S.\sqrt{(3\sqrt2-1)^2}=|3\sqrt2-1|=3\sqrt2-1,t.k.3\sqrt2-1>0\\\\\sqrt{(3\sqrt2+1)^2}=|3\sqrt2+1|=3\sqrt2+1,t.k.3\sqrt2+1>0" align="absmiddle" class="latex-formula">