Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ?

Use the identity $$(x+y)^3=x^3+3x^2y+3xy^2+y^3= x^3+y^3+3xy(x+y).$$

Gauss gave a simple algorithm to rewrite a symmetric polynomial $f(x,y)$ as polynomial in the elementary symmetric polynomials $\,s_1\! = x+y,\ \ s_2 = xy.$ Namely if $f$ has highest degree term $\ c x^a y^b $ in the lex (dictionary) order (i.e. $\,(a,b) > (c,d)\, $ if $\,a >c,\,$ or $\,a= c\,$ and $\, b > d)\,$ then cancel the highest term of $\,f\,$ by subtracting $\,cs_1^{a-b} s_2^b,\, $ then recurse on what remains.

Let's perform Gauss's algorithm on the example at hand $\, f = x^3 + y^3.\, $ Since $\,(3,0) > (0,3)\,$ the highest degree monomial is $\ 1\cdot x^\color{#0a0}3y^\color{#c00}0,\, $ so we subtract $\ 1\cdot s_1^{\color{#0a0}3-\color{#c00}0} s_2^\color{#c00}0\, =\, (x+y)^3 $ yielding

$$\ x^3+y^3\ -\ (x+y)^3\, =\ {-}3x^2 y - 3x y^2$$

By $(2,1)>(1,2),\, $ RHS has high term $\,-3x^{\color{#0a0}2} y^\color{#c00}1$ so we subtract $\, {-}3 s_1^{\color{#0a0}2-\color{#c00}1} s_2^\color{#c00}1 =\, -3(x\!+\!y)(xy)$

$$\ x^3+y^3\, -\ (x+y)^3\, +\ 3(x+y)(xy) \ =\ 0$$

So the algorithm terminates, yielding $\ f = s_1^3 - 3s_1 s_2.\ $ Since you know the values $\,s_1 = x+y,\ $ and $\ s_2 = xy,\ $ you can now calculate $f$ by using the prior equation.

This same algorithm works for polynomials in any number of variables. It reduces such problems to rote mechanical computation, i.e. no guesswork is required to solve such problems, only simple polynomial arithmetic. The algorithm yields a constructive interpretation of the Fundamental Theorem of Symmetric Polynomials, that every symmetric polynomial has a unique representation as a polynomial in the elementary symmetric polynomials.

Gauss's algorithm may be viewed as a special case of Gröbner basis methods (which may be viewed both as a multivariate generalization of the (Euclidean) polynomial division algorithm, as well as a nonlinear genralization of Gaussian elimination for linear systems of equation). Gauss's algorithm is the earliest known use of such a lexicographic order for term-rewriting (now mechanized by the Grobner basis algorithm and related methods).

Hrm. How can we make $x^3 + y^3$ out of that?

What tools do we have available to us? We have $x+y$. We have $xy$. And we have arithmetic.

I don't know yet how to make $x^3 + y^3$ exactly. Let's focus on one part: let's make $x^3$, and then see what we can do with what's left over.

It's pretty easy to make $x^3$ by multiplying things together. We could take $(xy) \cdot (xy) \cdot (xy)$. We could take $(x+y) \cdot(x+y) \cdot (xy)$. There are others. Let's experiment and actually try all of them. Multiplying them out, we can make

• $x^3 y^3$
• $x^3 y^2 + x^2 y^3$
• $x^3 y + 2 x^2 y^2 + x y^3$
• $x^3 + 3 x^2 y + 3 x y^2 + y^3$

That last one looks intriguing: it has $x^3$ all by itself without other terms multiplied in. It even has $y^3$ in it too.

So let's go with that, and try to subtract off the rest. Now, the question is

How do we make $3x^2 y + 3 x y^2$

If we can solve that, then we can combine that answer with the above to make $x^3 + y^3$.

Its simple. Use the identity ${x^3 + y^3 =(x+y)(x^2 +y^2 -xy) }$

In the above identity; make modifications and bring it to this form : ${x^3+y^3 = (x+y)((x+y)^2 -3xy) }$

Now putting the values of ${x+y}$ and ${xy}$ in the above equations, you should get the correct answer.

:)