$(4 + 2 i)^{3} = 88 i + 16$

Explanation

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$(4 + 2 i)^{3} \overset{◯}{1} 64 + 96 i + 48 i^{2} + 8 i^{3} \overset{◯}{2} \overset{◯}{3} 64 + 96 i - 48 - 8 i \overset{◯}{4} 88 i + 16$
①	Find $(4 + 2 i)^{3}$ using formula $(A + B) = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$ where $A = 4$ and $B = 2 i$ . $(4 + 2 i)^{3} = 4^{3} + 3 \cdot 4^{2} \cdot 2 i + 3 \cdot 4 \cdot (2 i)^{2} + (2 i)^{3} = 64 + 96 i + 48 i^{2} + 8 i^{3}$
②	$48 i^{2} = 48 \cdot (- 1) = - 48$
③	$8 i^{3} = 8 \cdot i^{2} \cdot i = 8 \cdot (- 1) \cdot i = - 8 \cdot i$
④	Combine like terms: $96 i - 8 i - 48 + 64 = 88 i + 16$

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(4+2i)3=88i+16

$(4 + 2 i)^{3} = 88 i + 16$