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

 Jun 6, 2024
 #1
avatar+806 
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We get two seperate equations from the problem. 

 

We have

\( a^3 + 3ab^2 = 679 \\ 3a^3 - ab^2 = 615\)

 

Let's multiply the second equation by 3 so that we can get 3a^3 in both equations

\(3a^3+9a^2=2037\)

\(3a^3-ab^2=615\)

 

Now, subtract the second equation from the first equation. We get

\(10ab^2 = 1422 \\ ab^2=142.2\)

 

Now, sub this value back into the first equation to get that

\(a^3+3(142.2)=679 \\ a^3+426.6=679 \\ 252.4=a^3 \\ a=\sqrt[3]{252.4} \\ a \approx 6.19\)

 

Now we have a, we can find b. 

\(b^2 = (142.2)/(6.19) \\ b^2 \approx 22.97 \\ b \approx \sqrt{22.97} \\ b \approx 4.79\)

 

So, we have

 \(a - b \approx 6.19 - 4.79 \\ a - b \approx 1.40\)

 

So 1.4 is our answer

 

Thanks! :)

 Jun 6, 2024

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