11.17.11
Aim: Solving Cubics of the form x³ = fx + g

Do Now: Solve by 'guess and check.'
1. x³ = 12x + 65
x³ - 12x = 65
x(x²-12) = 65
x = 5
2. x³ = 6x + 9
x³ - 6x = 9
x(x² - 6) = 9
x = 3
3. x³ = 6x + 40
x³ - 6x = 40
x(x² -6) = 40
x = 4
4. x³ -15x = 70
x(x² - 15) = 70
*IRRATIONAL* about 5.3

1515: Del Ferro figures out a process for solving certain cubic equations, depressed cubics without and x^2 term, are solved using his formula:
x³ = fx + g
1545: Cardano, who essentially invented probability, figures out the formula for solving depressed cubics:
given x³ = cx + d...
x =∛d/2 + √(d/2)² - (c/3)³ + ∛d/2 - √(d/2)² - (c/3)³
Another way to solve depressed cubics is by using the simultaneous equation method:
given x = a + b
x³ = (a + b)³, which is equal to....
x³ = a³ + 3a²b + 3ab² + b³
3a²b + 3ab² is equal to 3ab(a+b)
so...if x = a + b, we can substitute x into the equation:
3ab(a + b) becomes 3ab(x)
Our new equation is now equal to:
x³ = 3abx + a³ + b³

We can now solve the depressed cubic 6x + 40 using this equation:
x³= 6x + 40
if x³ = 3abx + a³ + b³,
-then 3ab = 6,
-ab=2
-and b = 2/a
this also means that
- a³ + b³ = 40
- if we substitute b for (2/a) in the equation above, we get:
a³ + (2/a)³ = 40
or
a³ + 8/a³ = 40
- if we multiply everything by a³ so as that the fraction cancels out, we get:
a⁶ + 8 = 40a³
or
a⁶ - 40a³ + 8 = 0
- this is equal to (a³)² - 40a³ + 8 = 0, where a³ is treated like x.
- Because this is now a quadratic equation, we can solve by using the quadratic formula and instead of solving for x, we can solve for a³.
- Eventually, we get that a³ = 20 + √392, which means that
a = ∛20 + √392
and
b = ∛20 - √392