How do I compute the sum of 2 squares

Question

if $x+y=a$ and $xy=b$, what does $x^3+y^3$ equal? I understand that $x^3+y^3=(x+y)(x^2-xy+y^2)$ but I don't see how I can figure out what $x^2$ or $y^2$ equals

Answer 1

Begin with finding a formula for $(x+y)^2$ and the rest is obvious.

Answer 2

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

Answer 3

You don't really need to worry about the squares. Instead, start by expanding $(x+y)^3$ and then collect terms:

$$(x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y)$$

Answer 4

You know $x+y$ and $xy$, and you want to find $x^2-xy+y^2$.

The trinomial should make you think of $a^2=(x+y)^2=x^2+2xy+y^2$. The difference of the two is $3xy=3b$, so that $x^3+y^3=a.(a^2-3b)$.

In case you don't want to take the shortcut, you can work by solving for $x$ and $y$: $y=a-x$ and $xy=b=x(a-x)$, so that $x^2-ax+b=0$, giving $x=\frac{a+\sqrt\delta}{2}$, and $y=\frac{a-\sqrt\delta}{2}$ (where $\delta=a^2-4b$).

Using the binomial expansion, $x^3+y^3=\frac{a^3+3a^2\sqrt\delta+3a\sqrt\delta^2+\sqrt\delta^3}{8}+\frac{a^3-3a^2\sqrt\delta+3a\sqrt\delta^2-\sqrt\delta^3}{8}=\frac{a^3+3a\delta}{4}=a^3-3ab$.