Solve this

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.

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