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 Math 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 y  - _______ ³Ö -x + Ö y
 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 )³
 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