Simplify the given complex expression: ${i^{18}} - 3{i^7} + {i^2}(1 + {i^4}){( - i)^{26}}$.

Hint: According to given in the question we have to find the value or simplify the given complex expression but first of all we have understand about complex expression as explained below:
Complex number: A complex number as a number that can be written in the form of a + ib, where i is the imaginary unit and a and b are real numbers. On multiplying complex numbers, it’s useful to remember that the properties we use when performing arithmetic with real numbers work similar for complex numbers. Sometimes, thinking of $i$ as a variable, like a, is helpful. Then with just a few adjustments at the end, we can multiply just as we expect.
And to solve the complex expressions as we know that,
$
   \Rightarrow {i^2} = - 1 \\
   \Rightarrow {i^4} = 1 \\
 $
Hence, to solve the complex expression first of all we will try to make the imaginary term i in form of ${i^2}$ or ${i^4}$ so that we can place the value of ${i^2}$= -1 or ${i^4}= 1$ and can find the value of the given complex expression.

Complete step by step answer:
Step 1: First of all we have to try to make the given imaginary term i in the complex expression in form of ${i^2}$ or ${i^4}$
$ \Rightarrow {({i^2})^9} - 3i \times {({i^2})^3} + {i^2}[1 + {({i^2})^2}]{( - {i^2})^{13}}$…………………………(1)
Step 2: Now, to solve the obtained expression we have to place the value of ${i^2}$ as mentioned in the solution hint.
$ \Rightarrow {( - 1)^9} - 3i \times {( - 1)^3} + ( - 1)[1 + {( - 1)^2}]{(1)^{13}}$…………………….(2)
Step 3: On solving the expression (2) as obtained in the step 2
$
   = - 1 + 3i + 2 \\
   = 1 + 3i \\
 $

Hence, by arranging the terms of the given complex expression and placing the value of i we have simplify the expression ${i^{18}} - 3{i^7} + {i^2}(1 + {i^4}){( - i)^{26}} = 1 + 3i$

Note:
Thinking of $i$ as a variable, like a, is helpful. Then with just a few adjustments at the end, we can multiply just as we expect.
Multiplying to imaginary (but not complex)numbers together works in the similar way.