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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 24, 2024
 #1
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First, let's express a^3 in terms of ab^2. We have

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

 

Now, let's subsitute this value for a into the second equation. We get that

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

 

Now, we simply solve for ab^2. We have that

\(\\ 2037 - 9ab^2 - ab^2 = 615 \\ \\9ab^2 + ab^2 = 2037 - 615 \\ 10ab^2 = 1422 \\ ab^2 = (1422)/(10) \\ ab^2 = 142.2 \\\)

Subsituting this value back into the first equation, we have

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

Now, we solve for b. We get

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

 

Thus, 

\( a - b \approx 1.40 \)

 

So our answer is approximately 1.4

Thanks! :)

 Jun 24, 2024

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