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Question

Answer

$(2 i + 2)^{4} = - 64$

Explanation

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$(2 i + 2)^{4} \overset{◯}{1} \overset{◯}{2} \overset{◯}{3} \overset{◯}{4} 16 i^{4} + 64 i^{3} + 96 i^{2} + 64 i + 16 \overset{◯}{5} \overset{◯}{6} \overset{◯}{7} 16 - 64 i - 96 + 64 i + 16 \overset{◯}{8} - 64$
①	$(2 i + 2)^{4} = (2 i + 2)^{2} \cdot (2 i + 2)^{2}$
②	Find $(2 i + 2)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = 2 i$ and $B = 2$ . $(2 i + 2)^{2} = (2 i)^{2} + 2 \cdot 2 i \cdot 2 + 2^{2} = 4 i^{2} + 8 i + 4$
③	Multiply each term of $(4 i^{2} + 8 i + 4)$ by each term in $(4 i^{2} + 8 i + 4)$ . $(4 i^{2} + 8 i + 4) \cdot (4 i^{2} + 8 i + 4) = 16 i^{4} + 32 i^{3} + 16 i^{2} + 32 i^{3} + 64 i^{2} + 32 i + 16 i^{2} + 32 i + 16$
④	Combine like terms: $16 i^{4} + 32 i^{3} + 16 i^{2} + 32 i^{3} + 64 i^{2} + 32 i + 16 i^{2} + 32 i + 16 = = 16 i^{4} + 64 i^{3} + 96 i^{2} + 64 i + 16$
⑤	$16 i^{4} = 16 \cdot i^{2} \cdot i^{2} = 16 \cdot (- 1) \cdot (- 1) = 16$
⑥	$64 i^{3} = 64 \cdot i^{2} \cdot i = 64 \cdot (- 1) \cdot i = - 64 \cdot i$
⑦	$96 i^{2} = 96 \cdot (- 1) = - 96$
⑧	Combine like terms: $- 64 i + 64 i - 96 + 16 + 16 = - 64$

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