Finally, let’s check (−1/2,−1/4) in that equation:
2 x^3 −y= 0
2 × (−1/2)^3 − (−1/4)= 0
2 × (−1/8)+ 1/4 = 0
−1/4+ 1/4 = 0
0 = 0
- Here are the two equations in their original forms:
4 x^3 + 2 x^2 + 2 x− 2 y− 8 = 0
and
3 x^3 − 2 x^2 + 4 x−y− 5 = 0
The first of these can be simplified if we divide through by 2. That gives us
2 x^3 +x^2 +x−y− 4 = 0
We can morph this into a function of x by adding y to each side, getting
y= 2 x^3 +x^2 +x− 4
The second original equation can also be modified by adding y to each side, obtaining
the function
y= 3 x^3 − 2 x^2 + 4 x− 5
Mixing the right sides of these two cubic functions, we get
2 x^3 +x^2 +x− 4 = 3 x^3 − 2 x^2 + 4 x− 5
Now let’s add the quantity (− 2 x^3 −x^2 −x+ 4) to each side and then transpose the equa-
tion left-to-right. That gives us
x^3 − 3 x^2 + 3 x− 1 = 0
The coefficients and constant in this equation show a certain symmetry. Whenever we see
a pattern of this sort in a cubic or higher-degree equation, it suggests that the equation
can be factored. After a few trials and errors, we discover that we have a binomial cubed:
(x− 1)^3 = 0
Which can be written out fully as
(x− 1)(x− 1)(x− 1) = 0
Chapter 27 707
