Question:
$\left\{i^{18}+\frac{1}{i^{25}}\right\}^{3}=2(1-i)$
Solution:
L.H.S $=\left\{i^{18}+\frac{1}{i^{25}}\right\}^{3}$
$\Rightarrow\left\{i^{4 \times 4+2}+i^{-4 \times 7+3}\right\}^{3}$
Since $i^{4 n}=1$
$i^{4 n+1}=i$
$i^{4 n+2}=-1$
$i^{4 n+3}=-1$
$=\left\{i^{2}+i^{3}\right\}^{3} .$
$=(-1-i)^{3}$
Applying the formula $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
We have,
$\left.+3 \mathrm{i}^{2}+3 \mathrm{i}+1\right)$
$i+3-3 i-1$
$=2(1-i)$
L.H.S = R.H.S
Hence proved.