Given x + y and x ⋅ y, what is x 3 + y 3 ?

Question

Given   x  +  y and   x  ⋅  y, what is       x    3    +      y    3  ?

Dwayne James · Accepted Answer

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} = x y .

. 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

c s_{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^{3} y^{0},

, so we subtract

1 \cdot s_{1}^{3 - 0} s_{2}^{0} = (x + y)^{3}

yielding

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

So the algorithm terminates, yielding

f = s_{1}^{3} - 3 s_{1} s_{2} .

. Since you know the values

s_{1} = x + y,

, and

s_{2} = x y,

, 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).

rigliztetbf · Accepted Answer

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

Given x + y and

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