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Let $a$ and $b$ be complex numbers. If $a + b = 4$ and $a^2 + b^2 = 6,$ then what is $a^3 + b^3?$

 Jun 21, 2024

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 #1
avatar+42 
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\((a+b)^2=a^2+b^2+2ab=6+2ab=16\), so ab=5. If we do \((a+b)^3\), we can expand it to \(a^3+a^2b+ab^2+b^3 \) via the binomial theorem. If we focus on the middle terms, we notice we can turn it into \(ab(a+b) \), which is equal to 5*4=20. Therefore, \(a^3+b^3=64-20=44 \).

 

Feel free to tell me if I did anything wrong! :D

 Jun 21, 2024
 #1
avatar+42 
+1
Best Answer

\((a+b)^2=a^2+b^2+2ab=6+2ab=16\), so ab=5. If we do \((a+b)^3\), we can expand it to \(a^3+a^2b+ab^2+b^3 \) via the binomial theorem. If we focus on the middle terms, we notice we can turn it into \(ab(a+b) \), which is equal to 5*4=20. Therefore, \(a^3+b^3=64-20=44 \).

 

Feel free to tell me if I did anything wrong! :D

Tottenham10 Jun 21, 2024

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