Question

Evaluate $(1+i\sqrt{2}{)}^8+(1-i\sqrt{2}{)}^8$

Answer 1

Consider the given equation,

${(1+i\sqrt{2})}^8+{(1-i\sqrt{2})}^8=-2^8$

L.H.S.,

${(1+i\sqrt{2})}^8+{(1-i\sqrt{3})}^8={\left[{(1+i\sqrt{2})}^2\right]}^4+{\left[{(1-i\sqrt{3})}^2\right]}^4$

$={\left[{(1^2+i^2{\sqrt{2}}^2+\mathrm{2.1.}i\sqrt{2})}^2\right]}^4+{\left[{(1^2+i^2{\sqrt{3}}^2-\mathrm{2.1.}i\sqrt{3})}^2\right]}^4$

$\because i=-1^2$

$={\left[(1+(-1)2+\mathrm{2..}i\sqrt{2})\right]}^4+{\left[(1+(-1)3-2i\sqrt{3})\right]}^4$

$={\left[(-1+2.i\sqrt{2})\right]}^4+{\left[(-2-2i\sqrt{3})\right]}^4$

$={\left[{(-1+2.i\sqrt{2})}^2\right]}^2+{\left[{(-2-2i\sqrt{2})}^2\right]}^2$

$={({(-1)}^2+2^2.i^2{\sqrt{2}}^2-4\sqrt{2})}^2+{({(-2)}^2+{(-2i\sqrt{3})}^2+8i\sqrt{3})}^2$

$\because i^2=-1$

$={(1-8-4\sqrt{2})}^2+{(4-12+8i\sqrt{3})}^2={(-7-4i\sqrt{2})}^2+{(-8+8i\sqrt{3})}^2$

$=49+\left(32i^2\right)+28i\sqrt{2}+64+384i^2-128\sqrt{3}$

$=113+416.i^2+28\sqrt{2}.i-128\sqrt{3}.i$

$=113+416.i^2+28\sqrt{2}.i-128\sqrt{3}.i$

$\because i^2=-1$

$=113+416(-1)+28\sqrt{2}.i-128\sqrt{3}.i$

$=306+28\sqrt{2}.i-128\sqrt{3}.i$