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Let $a$ and $b$ be real numbers such that $a^3 + 3ab^2 = 679$ and $3a^2 b + b^3 = 615.$ Find $a - b.$

 

Thanks! 

 Jun 2, 2023
 #1
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We square the first equation and the second equation to get \begin{align*} a^6 + 6a^4 b^2 + 9ab^4 &= 45761, \ 9a^4 b^2 + 6a^2 b^4 + b^6 &= 37225. \end{align*} Subtracting, we get a6−b6=8536 and hence (a3−b3)2=8536, or (a−b)(a2+ab+b2)=29. Since a and b are real, a2+ab+b2≥0. Hence, (a−b)≤29. But a3−b3=(a−b)(a2+ab+b2) is negative, so a−b must be positive. Hence a−b=29.

 Jun 2, 2023
 #4
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Thanks!

Guest Jun 7, 2023
 #2
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\(a^{3} + 3ab^{2}=679, \\ 3a^{2}b+b^{3}=615.\)

Subtract the second equation from the first,

\(a^{3}-3a^{2}b+3ab^{2}-b^{3}=64, \\ (a-b)^{3}=64, \\ a-b = 4.\)

 Jun 2, 2023
 #3
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That answer is wrong.

Guest Jun 6, 2023

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