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Q.
Simplest form of $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3$ is
Complex Numbers and Quadratic Equations
Solution:
$\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3=\left[\left(i^2\right)^9+\frac{1}{i^{25}}\right]^3$
$=\left[\left(i^2\right)^9+\frac{1}{i \cdot i^{21}}\right]^3$
$=\left[(-1)^9+\frac{1}{i\left(i^4\right)^6}\right]^3 \left(\because i^2=-1\right)$
$=\left[-1+\frac{1}{i}\right]^3 \left(\because i^4=1\right)$
$=\left[-1+\frac{1}{i} \times \frac{i}{i}\right]^3=\left[-1+\frac{i}{i^2}\right]^3$
$=[-1-i]^3$
$\left(\because i^2=-1\right)$
$=(-1)^3[1+i]^3$
$=-\left[(1)^3+i^3+3 \times 1 \times i(1+i)\right]$
$\left[\because(a+b)^3=a^3+b^3+3 a b(a+b)\right]$
$=-\left[1-i+3 i+3 i^2\right] \left(\because i^3=-i\right)$
$=-[1-i+3 i-3] \left(\because i^2=-1\right)$
$=-[-2+2 i]=2-2 i$
Students are suggested not to open the cube of expression at initial stage because the expression becomes complex by doing so.