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 Skip Navigation LinksMath Help > Basic Algebra > Factoring > Solving cubic, quartic, higher degree > Cubic Continued

The solution to a cubic equation (click here for a full explanation) has the following form:

    ______
³√x + Ö
    _______
³Ö -x + Ö
 
All the solutions produced by the cubic formula have this form, even if the solution is an integer.  For example,
    _________
³√10 + Ö 108 
    __________
³Ö -10 + Ö 108 
 
 = 2
 
I used trial and error to find dozens of integer-valued expressions of this form, and found they all have the following general form:
 
    _________________
³√3rp - 8r³ + pÖ p-3r² 
    __________________
³Ö -3rp + 8r³ + pÖ p-3r² 
 
 = 2r
where r is any real number, and p³3r²

 

 
Now I will prove this in two cases, Case I, where p=3r², and Case II, where p>3r².

CASE I, when p=3r²

    _______
³√3rp - 8r³ - 
    ________
³Ö -3rp+ 8r³
 
 = 
    __________
³√3r(3r²)- 8r³ - 
    __________
³Ö -3r(3r²)+8r³
 
 = 
    _______
³√9r³- 8r³ - 
    ________
³Ö -9r³+8r³
 
 = 
    __
³√r³ - 
    ___
³Ö -r³
 
 = r-(-r) = 2r 
 

CASE II, when p>3r²

 
let 2w = 
    _________________
³√3rp - 8r³ + pÖ p-3r² 
    __________________
³Ö -3rp + 8r³ + pÖ p-3r² 
 
  
 
For simplicity in manipulating this equation,

let x = 3rp - 8r³, and

 
let y = p
  ______
Ö p-3r² 

In this proof we will use the following facts about x and y:

x2 = 64r6 - 48r2p + 9r2p2, and
y2 = -3r2p2 + p3.

Subtracting one from the other, we get

x2 - y2 = 64r6 - 48r2p + 12r2p2 - p3
x2 - y2 = (4r2 - p)3

Substituting x and y in our original equation, we now have,

 
2w = 
    ____
³√x + y  - 
    _____
³Ö -x + y 
 
  
 
8w³ = ( 
    ____
³√x + y  - 
    _____
³Ö -x + y 
 
 
8w ³ = (x+y) - 3
    __________
³√(x+y)²(-x+y) + 3
    __________
³√(x+y)(-x+y)² - 
 
(-x+y)
 
8w ³ = 2x - 3
    ___________
³√(-x²+y²)(x+y) + 3
    ____________
³√(-x²+y²)(-x+y) 
 
 
8w ³ = 2x - 3
    _______
³√(-x²+y²)  (
    ____
³√(x+y) - 
    _____
³√(-x+y) 
 
)
 
8w³ = 2x - 3(p-4r²) (2w)
4w³ = x - 3(p-4r²) (w)
4w³ = 3rp - 8r ³ - 3(p-4r²) (w)
4w³ = 3rp - 8r ³ - 3wp + 12r²w
4w³ = 3rp - 3wp - 8r ³ + 12r²w
0 = 3rp - 3wp - 8r³ + 12r²w - 4w ³
0 = 3rp - 3wp - (8r³ - 12r²w  - 4rw² + 4rw²+ 4w³)
0 = 3rp - 3wp - (8r³ - 4r²w - 4rw²) - (-8r²w + 4rw² + 4w³)
0 = 3rp - 3wp - r(8r² - 4rw - 4w²) + w(8r² - 4rw - 4w²)
0 = (r-w)(3p) - (r-w)(8r² - 4rw - 4w²)
0 = (r-w)(3p - (8r² - 4rw - 4w²))

Now we can see that either r=w (which is the thing we're trying to prove) or else
3p = 8r² - 4rw - 4w².  So if 3p is not equal to 8r² - 4rw - 4w², then r=w, and the proof
is complete.

(r+2w)² >= 0
r² + 4rw + 4w² >= 0
r² >= - 4rw - 4w²
9r² >= 8r² - 4rw - 4w²

Remember that p>3r², from the assumption of Case II, above, so 3p>9r².

3p > 9r² >= 8r² - 4rw - 4w²
3p >  8r² - 4rw - 4w²

So 3p is not equal to 8r² - 4rw - 4w²,
proving that r=w, or in other words,

    _________________
³√3rp - 8r³ + pÖ p-3r² 
    __________________
³Ö -3rp + 8r³ + pÖ p-3r² 
 
 = 2r
where r is any real number, and p³3r²

Related pages in this website

Quadratic formula

 

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