$(4 + 5 i)^{3} = 115 i - 236$

Explanation

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$(4 + 5 i)^{3} \overset{◯}{1} 64 + 240 i + 300 i^{2} + 125 i^{3} \overset{◯}{2} \overset{◯}{3} 64 + 240 i - 300 - 125 i \overset{◯}{4} 115 i - 236$
①	Find $(4 + 5 i)^{3}$ using formula $(A + B) = A^{3} + 3 A^{2} B + 3 A B^{2} + B^{3}$ where $A = 4$ and $B = 5 i$ . $(4 + 5 i)^{3} = 4^{3} + 3 \cdot 4^{2} \cdot 5 i + 3 \cdot 4 \cdot (5 i)^{2} + (5 i)^{3} = 64 + 240 i + 300 i^{2} + 125 i^{3}$
②	$300 i^{2} = 300 \cdot (- 1) = - 300$
③	$125 i^{3} = 125 \cdot i^{2} \cdot i = 125 \cdot (- 1) \cdot i = - 125 \cdot i$
④	Combine like terms: $240 i - 125 i - 300 + 64 = 115 i - 236$

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(4+5i)3=115i−236

$(4 + 5 i)^{3} = 115 i - 236$