Question

Stie careva? o fi usor dar nu am solutie cum sa fac:
$( \frac{1}{2+i} + \frac{1}{2-i} ) ^{3}$

Answer 1

Vom amplifica mai intai fractiile cu conjugatul numitorilor, folosindu-ne de aceasta formula:

$(a+bi)(a-bi)=(a)^2-(bi)^2=a^2-b^2i^2=a^2-b^2(-1)=a^2+b^2$

$(\frac{1}{2+i}+\frac{1}{2-i})^3=(\frac{2-i}{(2+i)(2-i)}+\frac{2+i}{(2+i)(2-i)})^3=(\frac{4}{2^2+1^2})^3=(\frac{4}{5})^3=\frac{64}{125}$