Evaluate: left[i^{18}+left(frac{1}{i}right)^{ | vedclass

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Question

(A) We have the expression $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$.First,simplify $i^{18}$ and $\left(\frac{1}{i}\right)^{25}$:$i^{18}

Vedclass · Accepted Answer

$2-2i$

Vedclass · Answer

(A) We have the expression $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$.First,simplify $i^{18}$ and $\left(\frac{1}{i}\right)^{25}$:$i^{18} = (i^4)^4 \cdot i^2 = 1^4 \cdot (-1) = -1$.$\left(\frac{1}{i}\right)^{25} = \frac{1}{i^{25}} = \frac{1}{(i^4)^6 \cdot i} = \frac{1}{1^6 \cdot i} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$.Substitute these values back into the expression:$[-1 + (-i)]^3 = [-(1+i)]^3 = (-1)^3 (1+i)^3$.Expand $(1+i)^3$ using the formula $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:$(1+i)^3 = 1^3 + i^3 + 3(1)(i)(1+i) = 1 - i + 3i(1+i) = 1 - i + 3i + 3i^2 = 1 + 2i - 3 = -2 + 2i$.Finally,multiply by $(-1)^3 = -1$:$-1 \cdot (-2 + 2i) = 2 - 2i$.

Evaluate: $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}$.

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