I approached it by attempting to factor it and then show that one factor can't be one and the other can't be prime. This gets nowhere, as you can't factor this expression. Is there another way to do it?
Show that $a^3+a^2b+ab^2+b^3$ is not a prime, if a and b are positive integers
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$\begingroup$
algebra-precalculus
prime-numbers
factoring
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4Hint: the expression is $0$ if $a=-b$, so it has a factor of $a+b$. – 2017-01-26
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0@dxiv not a hint really, but a solution .. +1 nonetheless – 2017-01-26
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0@windircurse Thanks. I consider a hint to be something that requires understanding and additional work before claiming as one's own solution. In that sense, both my comment and (hinted) answer qualify as hints. – 2017-01-26
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1Note that hints showing two factors really are hints, because for a solution you have to invoke the test for primality - what happens if one of the factors is $1$? And the question says nothing about numbers being positive (integers can, I think be assumed, but should be stated) to $\pm 1$ could be tested. – 2017-01-26
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0That *can* be a prime. For example, if `a = 3` and `b = -2`, then the result is 13 -- which is prime. – 2017-01-26
3 Answers
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$a^3+b^3+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+ab(a+b)\\=(a+b)(a^2-ab+b^2+ab)\\=(a+b)(a^2+b^2)$
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Hint: $a^3+a^2b+ab^2+b^3=(a^2 + b^2)(a + b)$.
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1Hint? More of a solution. – 2017-01-26
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1Not yet a solution. We need to argue that this is not $1\cdot p$ or $p\cdot 1$ for a prime. In addition, $a$ or $b$ could be negative. – 2017-01-26
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Hint: $(a-b)(a^3+a^2b+ab^2+b^3) = a^4-b^4=(a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a^2+b^2)$