Answer
$15+20i$
Work Step by Step
Note: the standard form of a complex number is $a+bi$ where $i = \sqrt -1$ and $a$ and $b$ are real numbers.
Note: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(2+i)^{3}(2-i)$
The complex number can be rewritten as:
$(2+i)^{3}(2-i) = (2+i)^{2}(2+i)(2-i)$
Solve for $(2+i)^{2}$:
$(2+i)^{2} = (2+i)(2+i)$
Expand out the complex number:
$(2+i)^{2} = 4+2i+2i+i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(2+i)^{2} = 4+2i+2i+(-1)$
Simplify:
$(2+i)^{2} = 4+2i+2i-1$
Combine like terms, to express the complex number in standard form:
$(2+i)^{2} = 3+4i$
Solve for $(2+i)(2-i)$:
Expand out the complex number:
$(2+i)(2-i) = 4 -2i + 2i - i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(2+i)(2-i) = 4 -2i + 2i -(-1)$
Simplify:
$(2+i)(2-i) = 4 -2i + 2i +1$
Combine like terms, to express the complex number in standard form:
$(2+i)(2-i) = 5$
Since $(2+i)^{2} = 3+4i$ and $(2+i)(2-i) = 5$, substitute into the expression for $(2+i)^{3}(2-i)$ from above:
$(2+i)^{3}(2-i) = (2+i)^{2}(2+i)(2-i)$
$(2+i)^{3}(2-i) = (3+4i)(5)$
Expand out the complex number:
$(2+i)^{3}(2-i) = 15+20i$
Therefore, the standard form of the complex number is:
$(2+i)^{3}(2-i) = 15+20i$
