Question Number 71406 by mind is power last updated on 15/Oct/19

Hello Solve (x^2 )^(1/3) −3((x(x−1)))^(1/3) +2((x−1))^(1/3) =0

Commented byPrithwish sen last updated on 15/Oct/19

x^2 +(x−1)−27x(x−1)+18x(x−1)^(2/3) =0 putting x =a^3 and x−1 = b^3 ⇒a^3 = 1+b^3 ∴ a^6 +8b^3 −27a^3 b^3 +18a^3 b^2 = 0 ⇒26b^6 −18b^5 +17b^3 −18b^2 −1=0 and I got a equation of 6 degrees. Sir please comment.

Commented byMJS last updated on 15/Oct/19

a^(1/3) −b^(1/3) +c^(1/3) =0 a^(1/3) +c^(1/3) =b^(1/3) a+c+3a^(1/3) c^(1/3) (a^(1/3) +c^(1/3) )=b 3a^(1/3) b^(1/3) c^(1/3) =b−a−c 27abc=(b−a−c)^3 a^3 −b^3 +c^3 +21abc−3(a^2 b+a^2 c+ab^2 +ac^2 +b^2 c−bc^2 )=0 a=x^2 b=27x(x−1) c=8(x−1) x^6 −((19203)/(4394))x^5 +((61719)/(8788))x^4 −((92387)/(17576))x^3 +((4299)/(2197))x^2 −((840)/(2197))x+((64)/(2197))=0 I found no exact solution x_1 ≈.172484 x_2 ≈1.72890 x_(3, 4) ≈.230494±.209231i x_(5, 6) ≈1.00395±.0115209i if we generally use (z)^(1/3) =abs (z)^(1/3) ×e^(i((arg (z))/3)) only x_2 is a solution if we use the common exception ((−r))^(1/3) =−(r)^(1/3) also x_1 is a solution the complex solutions are false

Commented byPrithwish sen last updated on 15/Oct/19

you just turn the equation in cubic form. Great sir.

Commented bymind is power last updated on 15/Oct/19

thank you !, i have no better solution my bee we can′t find exacte root ′ or its hard we need find function g (t) such that if we put x=g(t) equation withe t is easier i try but no result

Commented byMJS last updated on 15/Oct/19

your equation is equivalent to mine

Commented bymind is power last updated on 15/Oct/19

i have did like you sir !