Question

if alpha,beta are the roots of eqn ax²+ bx+c=0, then find the quadratic eqn whose roots are alpha³, beta ³​

Answer 1

ax^2+bx+c=0

alpha+beta = -b/a - (1)

alpha*Beta=c/a - (2)

now we do cubes on both side 0f  1

(alpha+beta)^3=(-b/a)^3

alpha^3+beta^3+3alpha*beta(Alpha+Beta)=(-b^3)/a^3

alpha^3+beta^3+3*c/a*-b/a=(-b^3)/a^3              {using 1 and 2}

alpha^3+beta^3-3bc/a^2=(-b^3)/a^3

alpha^3+Beta^3=(-b^3)/a^3+3bc/a^2

now we do cubes bhs of 2nd eq

(alpha*beta)^3=(c/a)^3

alpha^3beta^3=c^3/a^3

now in order to form a quadratic equation when roots are given we use formula

x^2+(alpha+beta)x+alpha*beta=0

now here alpha= alpha^3 and beta=beta^3

so u will get the equation

x^2+(alpha^3+beta^3)x+alpha^3*beta^3=0

we have already found the value of this in above u can put it and get your answer