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# System

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I'm very stuck on this system problem.

Let a and b be real numbers such that a^3 + 3ab^2 = 679 and a^3 - 3ab^2 = 673.  Find a - b.

Jan 22, 2023

#1
+118207
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I think it is possible that you have mistyped the question but what you HAVE asked is still quite doable.

Subtract the second from the first and you get

$$6ab^2=6\\ 3ab^2=3$$

So now I have

$$a^3+3=679\qquad and \qquad a^3-3=673\\ a^3=676\\ and\\ 3*\sqrt[3]{676}*b^2= 3\\ \sqrt[3]{676}*b^2= 1\\ b=\pm \sqrt{1-\sqrt[3]{676}}\\ a-b= \sqrt[3]{676} \pm \sqrt{1-\sqrt[3]{676}}\\$$

You need to check that, I could easily have made a careless mistake.

NOW check your question, i really do not think you wrote it down properly.

Jan 23, 2023