Evaluate $\left[i^{22}+\left(\dfrac{1}{i}\right)^{25}\right]^3$

Given the expression $\left[i^{22} + \left(\dfrac{1}{i}\right)^{25}\right]^3$.

First, simplify the powers of $i$. We know that $i^2 = -1$, $i^4 = 1$, and $\dfrac{1}{i} = -i$.

$i^{22} = (i^4)^5 \times i^2 = 1^5 \times (-1) = -1$.

$\left(\dfrac{1}{i}\right)^{25} = (-i)^{25} = (-1)^{25} \times i^{25} = -1 \times (i^4)^6 \times i = -1 \times 1^6 \times i = -i$.

Substituting these values into the expression:

$\left[-1 + (-i)\right]^3 = \left[-(1 + i)\right]^3 = -1 \times (1 + i)^3$.

Expanding $(1 + i)^3$ 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, multiplying by $-1$:

$-1 \times (-2 + 2i) = 2 - 2i = 2(1 - i)$.

Answer: $2(1-i)$