Newton identities (or Newton-Girard formulae)

Question

Given $x+y=a$, $\quad x^2 + y^2 = b$, how do you use Newton's identities to come up with the expression for $x$ cubed plus $y$ cubed.

I am getting $(1/2)a(3b-a)$ which is wrong when you test with numbers.

Answer 1

Since you're working with two variables, it would make sense to look at the identities for a quadratic polynomial.
I shall follow the notation used here.

First, we construct a polynomial that has $x$ and $y$ as roots. Using the fact that $x + y = a$, we get that
$$2xy = (x + y)^2 - (x^2 + y^2) = a^2 - b.$$

Thus, the monic polynomial with $x$ and $y$ as roots is

$$P(z) = x^2 - ax + \dfrac{1}{2}(a^2 - b).$$

Now, using Newton's formula gives us

$$x^3 + y^3 = - (-a)(x^2 + y^2) - \dfrac{1}{2}(a^2 - b)(x+ y)$$

This simplifies to $x^3 + y^3 = \boxed{\dfrac{1}{2}(3ab-a^3)}$.

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Note that the other answer is simpler and you should probably use that unless you're forced to use the Newton-Girard formulae.

Answer 2

$x^3+y^3=(x+y)(x^2-xy+y^2)=a(b-xy)$

So we just need to determine $xy$. since $x^2+y^2=b$, it follows that $x^2+2xy+y^2=b+2xy$ and so $$(x+y)^2=b+2xy$$ which means $a^2=b+2xy\implies xy=\frac{1}{2}(a^2-b)$ so finally $$x^3+y^3=a(b-xy)=a(b-\frac{1}{2}(a^2-b))=\frac{1}{2}a(3b-a^2)$$ Hope this helps!

Answer 3

So $x+y=a$ and $x^2+y^2=b$.

Compute $(x+y)^2=x^2+2xy+y^2$ so we may solve for $xy$ in terms of $a$ and $b$.

Compute $(x+y)^3=x^3+3x^2y+3xy^2+y^3$ which equals $(x^3+y^3)+3xy(x+y)$ so...