OK, so we need real coefficients. First at all we can take those two factors from my previous answer (a -b(1/2 - i sqr(3) /2) )( a - b ( 1/2 +i sqr(3) / 2)) and multiply them as (u + v)(u-v) = u^2 - v^2. So one gets (a- b/2)^2 - b^2 (-1) 3/4 = a^2 -ab + b^2/4 + b^2 (3/4) = a^2 - ab + b^2. The other factor remains a + b.

But I understand now that the question is much more about didactic of mathematics and that the pupils maybe do not know the complex numbers and the roots of unity. In this situation I would however start again with the polynomial X^3 + 1. I will observe that X = -1 is a root, so by Horner's Lemma (in Romanian School is called so) I would conclude that the polynomial X^3 + 1 is divisible by the polynomial X + 1. Then I would make polynomial division to get X^3 + 1 = (X+1)(X^2 - X + 1). Finally I would replace again X with a/b where a and b are new variables, and I would multiply all with b^3, to conclude that a^3 + b^3 = (a + b)(a^2 - ab + b^2). 

But, you know, I don't teach - so I don't know exactly what do they understand and what not. I am sure that I can explain them all of this, but I don't feel before how does it come to them. ...

Well, I don't really understand the difference between "to factor" and "to prove by multiplication", but I try. X^3 = -1 has the complex roots -1 x epsilon_i, where epsilon_i are the three roots of 1. So epsilon_i are 1, -1/2 + i sqr(3) / 2 and -1/2 - i sqr(3) / 2, and so X^3 + 1 = (X + 1) ( X - 1/2 + i sqr(3)/2 ) ( X - 1/2 - i sqr(3) / 2). Finally it follows that a^3 + b^3 = (a + b) (a -b(1/2 - bi sqr(3) /2) )( a - b ( 1/2 - i sqr(3) / 2)).

Dear Mihai!

Thank you very much for your answer and expressing of perplexity. During the formulation of the question, I felt that not all is expressed well. In Russian mathematics teaching, there exists a class of standard problems with the clue word translating into the English as "factor". Now I will try to improve my question by selection (or matching) a more appropriate word: "to split", "to decompose" into a product of polynomial factors. (And, of course, my fault was that I did not mention that decomposition is meant over the field of reals.) 

So, the improved formulation of my question is this:

• Decompose a^3 + b^3 into a product of polynomials with real coefficients

(of course, of polynomials of degree not less than one). Why I posed this question? In Russian mathematics teaching is quite ordinary the situation where a pupil is given the problem to decompose something, but when he do not solve it, he is given a "solution" in which teacher instead of true decomposition prove what is needed by multiplication.

OK, so we need real coefficients. First at all we can take those two factors from my previous answer (a -b(1/2 - i sqr(3) /2) )( a - b ( 1/2 +i sqr(3) / 2)) and multiply them as (u + v)(u-v) = u^2 - v^2. So one gets (a- b/2)^2 - b^2 (-1) 3/4 = a^2 -ab + b^2/4 + b^2 (3/4) = a^2 - ab + b^2. The other factor remains a + b.

But I understand now that the question is much more about didactic of mathematics and that the pupils maybe do not know the complex numbers and the roots of unity. In this situation I would however start again with the polynomial X^3 + 1. I will observe that X = -1 is a root, so by Horner's Lemma (in Romanian School is called so) I would conclude that the polynomial X^3 + 1 is divisible by the polynomial X + 1. Then I would make polynomial division to get X^3 + 1 = (X+1)(X^2 - X + 1). Finally I would replace again X with a/b where a and b are new variables, and I would multiply all with b^3, to conclude that a^3 + b^3 = (a + b)(a^2 - ab + b^2). 

But, you know, I don't teach - so I don't know exactly what do they understand and what not. I am sure that I can explain them all of this, but I don't feel before how does it come to them. ...

Dear Mihai,

Glad to see you again! You are my faithful (trusty) respondent. Thank you!

What about your answer, the problem is simpler, much simpler! Imagine the very beginning of the learning of algebra, something like identities of the shortened multiplication. They are proved, naturally, by multiplication, but using them one can "decompose by decomposition", that is, beginning with the given expression write down a sequence of identities ending with the desired product.

Thank you again!

Yours friendly and truthfully,

Aslanbek.

I think I know what do you mean. Maybe so:

a^3 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3 - 3a^2b - 3ab^2 = (a+b)^3 - 3ab(a+b) = (a+b)[(a+b)^2 - 3ab] = (a+b)[a^2 -ab + b^2]. I didn't think so before.

The point is that they have told us in the 7th class that for all n ODD, a^n + b^n = (a+b) (a^(n-1) - a^(n-2) b + .... + b^(n-1)), done by multiplication, of course. So this was kind of very basic knowledge about one does not think anymore. It is nice that a^n - b^n = (a-b) (a^(n-1) + a^(n-2) b + ... + b^(n-1)) is true for all n, not only ODD n. 

For the ODD n I find it more schocking that it follows that the only prime which is sum of two non-degenerate odd powers is 2. This is quite different from the squares, because there are infinitely many primes sum of two squares: all those which are 4k+1!

Yes, Mihai, just so!

Congratulations with the right solution and with your new discovery!

Sincerely!

Quite generally, we have:

i) an - bn = (a - b)·(an-1·b0 + an-2·b1  + an-3·b2 + … + a2·bn-3 + a1·bn-2 + a0·bn-1) = (a - b)·∑k=0,n-1 [a n-k-1·bk]; being n positive integer;

ii) an - bn = (a + b)·(an-1·b0 - an-2·b1 + an-3·b2 - … - a2·b n-3 + a1·bn-2 - a0·bn-1) = (a+b)·∑k=0,n-1 [(-1)k·an-k-1·bk]; being n even positive integer;

iii) an + bn = (a + b)·(an-1·b0 - an-2·b1 + an-3·b2 - … + a2·bn-3  -  a1·bn-2 + a0·bn-1) = (a+b)·∑k=0,n-1 [(-1)k·a n-k-1·bk], being n odd positive integer.

As a special case of last formula for n = 3: a3 + b3 = (a + b)·(a2 - a·b + b2).

Dear Carlos,

Had you read the statement of the problem? 

Dear Aslanbek:

I have tried to answer your question, which I have read, and also all answers given before, including your posts. I did not realise that you were posing a problem. I did not tried to prove or solve nothing.

Regards