Prove the cubic formula

Answer 1

Given ax³ + bx² + cx + d = 0 and a =/= 0, we have the following:

x³ + bx²/a + cx/a + d/a = 0, so x³ + 3(b/3a)x² + 3(b/3a)²x + cx/a + d/a = 3(b/3a)²x,

so x³ + 3(b/3a)x² + 3(b/3a)²x + (b/3a)³ + cx/a + d/a = 3(b/3a)²x + (b/3a)³,

so (x + b/3a)³ + cx/a + c(b/3a)/a + d/a = (b/3a)² (3x + b/3a) + c(b/3a)/a,

so (x+b/3a)³+(c/a)(x+b/3a)+d/a+(b/3a)²(2b/3a)=(b/3a)²(3x+b/3a)+(b/3a)²(2b/3a)+(c/a)(b/3a),

so (x+b/3a)³+(c/a)(x+b/3a)+d/a+(b/3a)²(2b/3a)=(b/3a)²(3x+3b/3a)+(c/a)(b/3a),

so (x+b/3a)³+(c/a)(x+b/3a)-3(b/3a)²(x+b/3a)+d/a+(b/3a)²(2b/3a)-(c/a)(b/3a)=0,

so (x+b/3a)³ + [c/a-3(b/3a)²] (x+b/3a) + d/a + 2(b/3a)³ - (bc/3a²) = 0,

so y³ + Ay + B = 0, with y = x+b/3a, A = c/a-3(b/3a)², and B = d/a+2(b/3a)³-(bc/3a²)

y = u + v, so y³ = (u+v)³ = u³ + 3u²v + 3uv² + v³ = u³ + v³ + 3uv(u+v) = u³ + v³ + 3uvy,

so y³ - (u³ + v³ + 3uvy) = 0, so y³ - (3uv)y - (u³ + v³) = 0 = y³ + Ay + B,

so A = -3uv and B = -(u³ + v³), so A³ = -27u³v³ and 27u³B = -27(u³)² - 27u³v³,

so 27u³B = -27(u³)² + A³, so 27(u³)² + 27B(u³) - A³ = 0, so now we have, from the

quadratic formula: u³ = { -(27B) +- ²√[(27B)² - 4*(27)*(-A³)] } / { 2*(27) },

so u³ = { -27B +- ²√[(27)²B² + 4*(27)²*A³/27] } / { 2*(27) },

so u³ = [ -B +- ²√(B² + 4*A³/27) ] / 2, so B = -([ -B +- ²√(B² + 4*A³/27) ]/2 + v³),

so v³ = -B - [ -B +- ²√(B² + 4*A³/27) ]/2, so u = ³√{[ -B +- ²√(B² + 4*A³/27) ] / 2}

and v = ³√{-B - [ -B +- ²√(B² + 4*A³/27) ]/2}, so y = u + v, which is the following:

y = ³√{[ -B +- ²√(B² + 4*A³/27) ] / 2} + ³√{-B - [ -B +- ²√(B² + 4*A³/27) ]/2}

so since y = x + b/3a, then we have x = y - b/3a, so:

x = ³√{[ -B +- ²√(B² + 4*A³/27) ] / 2} + ³√{-B - [ -B +- ²√(B² + 4*A³/27) ]/2} - b/3a

where A = c/a-3(b/3a)², and B = d/a+2(b/3a)³-(bc/3a²).