Find the roots `alpha, beta, gamma,` of `x^3-11 x^2+36 x-36=0if2/beta=1/alpha+1/gamma.`

Find the roots alpha,beta,gamma, of x^(3)-11x^(2)+36x-36=0 if (2)/(beta)=(1)/(alpha)+(1)/(gamma)

If alpha , beta , gamma are the roots of x^3 + 4x +1 =0 then the equation whose roots are ( alpha^2)/( beta + gamma ) , ( beta^2)/( gamma + alpha) , ( gamma^2)/ ( alpha + beta) is

If alpha , beta , gamma are the roots of x^3 -3x +1=0 then the equation whose roots are alpha - (1)/( beta gamma) , beta - (1)/( gamma alpha ) , gamma - (1)/( alpha beta ) is

If alpha , beta , gamma are the roots of x^3 -3x +1=0 then the equation whose roots are alpha - (1)/( beta gamma) , beta - (1)/( gamma alpha ) , gamma - (1)/( alpha beta ) is

If alpha, beta ,gamma are the roots of 2x^(3) - 5x^(2) + 3x - 1 = 0 then (1)/(alpha beta ) + (1)/(beta gamma) + (1)/(gamma alpha) =

if alpha, beta, gamma are roots of x^3-3x^2+3x+7=0 , then find (alpha-1)/(beta-1)+(beta-1)/(gamma-1)+(gamma-1)/(alpha-1)

If alpha, beta, gamma are roots of x^(3)-2x^(2)+7x-1=0 then the value of (alpha+beta-gamma) (beta+gamma-alpha) (alpha+gamma-beta)