Answer
$52+47i$
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: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(4+i)^{3}$
The complex number can be rewritten as:
$(4+i)^{3} = (4+i)^{2}(4+i)$
Solve for $(4+i)^{2}$:
$(4+i)^{2} = (4+i)(4+i)$
Expand out the complex number:
$(4+i)^{2} = 16+4i+4i+i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(4+i)^{2} = 16+4i+4i+(-1)$
Simplify:
$(4+i)^{2} = 16+4i+4i-1$
Combine like terms, to express the complex number in standard form:
$(4+i)^{2} = 15+8i$
Since $(4+i)^{2} = 15+8i$, substitute into the expression for $(4+i)^{3}$ from above:
$(4+i)^{3} = (4+i)^{2}(4+i)$
$(4+i)^{3} =(15+8i)(4+i)$
Expand out the complex number:
$(4+i)^{3} =60+15i+32i+8i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(4+i)^{3} =60+15i+32i+8(-1)$
Simplify:
$(4+i)^{3} =60+15i+32i-8$
Combine like terms, to express the complex number in standard form:
$(4+i)^{3} = 52+47i$
