148 CHAPTER 5 ALGEBRA
32 = (X+y)^2 =�+2xy+l,
and since.xy = 3, we have � + y^2 = 3. Consequently,(^3) ·3= (x+ y)(� +l) =.J +�y+.xy^2 +l = x^3 +l +.xy(x+y).
From this, we conclude that
which is rather surprising.
Incidentally, what if we really wanted to find out the values of x and y? Here is an
elegant way to do it. The equation x + y = 3 implies (x + y)^2 = 3^2 , or
�+2xy+l =9.
Since.xy = 3, we subtract 4.xy = 12 from this last equation, getting
x^2 - 2.xy+l = - (^3).
This is a perfect square, and taking square roots gives us
x -y= ±iV3.
This equation is particularly useful, since it is given that x + y = 3. Adding these
two equations immediately gives us x = ( (^3) ± iV3) /2, and subtracting them yields
y = ( 3 � iV3) /2. Our two solutions for (x,y) are
(3+iV3 3 -iV3) ( 3 -iV3 3+iV3)
. •
2 ' 2 ' 2 ' 2
The Factor Tactic
Multiplication rarely simplifies things. Instead, you should
Factor relentlessly.
The following are basic formulas that you learned in an algebra class. Make sure
that you know them actively, rather than passively. Notice how Formula 5.2.4 instantly
solves Example 5.2.1!
5.2.2 (x+y)^2 =x^2 +2.xy+y^2.
5.2.3 (x-yf =x2 _2xy+y^2.
5.2.4 (x+ y)^3 = x^3 + 3x^2 y + 3.xy^2 + y^3 = x^3 + y^3 + 3.xy(x+ y).
5.2.5 (x -y)^3 = � -3x^2 y + 3.xy^2 - y^3 = x^3 - y^3 - 3.xy(x -y).
5.2.6 � -y^2 = (x -y)(x+ y).
5.2.7 XZ -yn = (x -y)(XZ-^1 +XZ-^2 y+XZ-^3 y^2 + ... + yn-l) for all n.
5.2.8 XZ + yn = (x+ y)(XZ-^1 -XZ-2y+XZ-3y^2 - ... + yn-l) for all odd n (the terms of
the second factor alternate in sign).