Question

Find the value of:
1) $(\sqrt{3}+1{)}^2-(\sqrt{3}-1{)}^2$
2) $(2+\sqrt{5}{)}^5+(2-\sqrt{5}{)}^5$

Answer 1

Find the value of

$1\left)\right(\sqrt{3}+1{)}^2-(\sqrt{3}-1{)}^2$ using

$(\sqrt{3}+1+\sqrt{3}-1)(\sqrt{3}+1-\sqrt{3}+1)$ $[a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$=2\sqrt{3}\times 2$

$=4\sqrt{3}$

$2\left)\right(2+\sqrt{5}{)}^5+(2-\sqrt{5}{)}^5$

$=(2+\sqrt{5}{)}^2(2+\sqrt{5}{)}^2+(2+\sqrt{5})+(2-\sqrt{5}{)}^2(2-\sqrt{5}\left)\right(2-\sqrt{5})$

$=(4+5\sqrt{5})(4+4\sqrt{5}+5)(2+\sqrt{5})+(4-4\sqrt{5})+(4-4\sqrt{5}+5{)}^2(2-\sqrt{5})$

$=(9+4\sqrt{5}{)}^2(2+\sqrt{5})+(9-4\sqrt{5}{)}^2(2-\sqrt{5})$

$=(81+72\sqrt{5}+80)(2+\sqrt{5})+(81-72\sqrt{5}+80)(2-\sqrt{5})$

$=(161+72\sqrt{5})(2+\sqrt{5})+(161-72\sqrt{5})(2-\sqrt{5})$

$=(322+161\sqrt{5}+144\sqrt{5}+72\times 5)+(322-161\sqrt{5}-144\sqrt{5}+72\times 5)$

$=322+161\sqrt{5}+144\sqrt{5}+380+322-16\sqrt{5}-144\sqrt{5}+360$

$=1364$