Question

Prove that
$(\sqrt{3}+\sqrt{2}{)}^6+(\sqrt{3}-\sqrt{2}{)}^6=970$

Answer 1

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

${=}^6{C}_0{\left(\sqrt{3}\right)}^6{+}^6{C}_1{\left(\sqrt{3}\right)}^5\sqrt{2}{+}^6{C}_2{\left(\sqrt{3}\right)}^4{\left(\sqrt{2}\right)}^2{+}^6{C}_3{\left(\sqrt{3}\right)}^3{\left(\sqrt{2}\right)}^3{+}^6{C}_4{\left(\sqrt{3}\right)}^2{\left(\sqrt{2}\right)}^4{+}^6{C}_0\left(\sqrt{3}\right)$

${\left(\sqrt{2}\right)}^5{+}^6{C}_6{\left(\sqrt{2}\right)}^6{+}^6{C}_0{\left(\sqrt{3}\right)}^6{-}^6{C}_1{\left(\sqrt{3}\right)}^5\left(\sqrt{2}\right){+}^6{C}_2{\left(\sqrt{3}\right)}^4{\left(\sqrt{2}\right)}^2{-}^6$

$C_3{\left(\sqrt{3}\right)}^3{\left(\sqrt{2}\right)}^3{+}^6{C}_4{\left(\sqrt{3}\right)}^2{\left(\sqrt{2}\right)}^4{-}^6{C}_5\left(\sqrt{3}\right){\left(\sqrt{2}\right)}^5{+}^6{C}_6{\left(\sqrt{2}\right)}^6$

$=2\left({}^6{C}_0{\left(\sqrt{3}\right)}^6{+}^6{C}_2{\left(\sqrt{3}\right)}^4{\left(\sqrt{2}\right)}^2{+}^6{C}_4{\left(\sqrt{3}\right)}^2{\left(\sqrt{2}\right)}^4{+}^6{C}_6{\left(\sqrt{2}\right)}^6\right)$$=2(27+\frac{6\times 5}{2}\times 3\times 3\times 2+\frac{6\times 5}{2}\times 3\times 2\times 2+8)$

$=2(27+30\left(9\right)+30\left(6\right)+8)$

$=2(27+270+180+8)$

$=2(450+35)$

$=2\times 485=970$