Prove the identity x^3 + y^3 = (x + y)^3 - 3xy(x + y).

Step-by-step solution

1. 1

   Choose a direction for the proof. Proving an identity means transforming one side into the other; you may not move terms across the equals sign as if solving an equation. The right-hand side is the complicated one, so expand it and aim to reach $x^3+y^3$. That direction needs only expansion, no factoring insight.

2. 2

   Expand the cube of the binomial. Using the binomial expansion,

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

   The two middle coefficients are both 3, which is exactly why a correction term with a factor of 3 will be able to remove them.

3. 3

   Expand the correction term.

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

   This is term-for-term the pair of mixed terms produced by the cube.

4. 4

   Subtract and watch the mixed terms vanish.

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

   Both middle terms cancel exactly, which completes the proof: the right-hand side equals $x^3+y^3$, so the identity holds for every $x,y$.

5. 5

   Why the identity is worth remembering. It expresses $x^3+y^3$ using only the elementary symmetric quantities $s=x+y$ and $p=xy$, namely $x^3+y^3=s^3-3ps$. That is the standard route to $x^3+y^3$ when a problem gives you the sum and product of two roots but not the roots themselves.

6. 6

   Numeric spot check. With $x=2,\,y=3$: the left side is $8+27=35$; the right side is $5^3-3\cdot 6\cdot 5=125-90=35$. Equal, as the algebra predicts.