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x-3-x-c-0-let-x-t-p-t-q-t-p-t-p-t-q-t-q-3-t-p-t-q-3-t-q-t-p-t-p-t-q-c-0-let-t-q-3-t-p-1-0-4t-1-p-q-4t-4p-3p-q-1-4t-4q-1-p-3q-3p-q-1

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Question Number 144768 by ajfour last updated on 04/Jul/21
x^3 −x−c=0 let x=(√(t+p))+(√(t+q)) (t+p)(√(t+p))+(t+q)(√(t+q)) +3(t+p)(√(t+q))+3(t+q)(√(t+p)) −(√(t+p))−(√(t+q))−c=0 let (t+q)+3(t+p)−1=0 ⇒ 4t=1−(p+q) ⇒ 4t+4p=3p−q+1 4t+4q=1−(p−3q) (3p−q+1)^(3/2) +3(1−p+3q)(√(3p−q+1)) −4(√(3p−q+1))−8c=0 2x=(√(3p−q+1))+(√(3q−p+1)) let (√(3p−q+1))=−1 ⇒ q=3p ⇒ −1−3(1−p+3q)+4−8c=0 ⇒ p=−(c/3) 2x=−1+(√(1−((8c)/3))) x=−(1/2)+(√((1/4)−((2c)/3))) for c=(1/4) x=−(1/2)+(1/(2(√3)))
x3−x−c=0letx=t+p+t+q(t+p)t+p+(t+q)t+q+3(t+p)t+q+3(t+q)t+p−t+p−t+q−c=0let(t+q)+3(t+p)−1=0⇒4t=1−(p+q)⇒4t+4p=3p−q+14t+4q=1−(p−3q)(3p−q+1)3/2+3(1−p+3q)3p−q+1−43p−q+1−8c=02x=3p−q+1+3q−p+1let3p−q+1=−1⇒q=3p⇒−1−3(1−p+3q)+4−8c=0⇒p=−c32x=−1+1−8c3x=−12+14−2c3forc=14x=−12+123

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