If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily, we get a new equation by putting the (m+d), =>m^3+d^3+3m*d^2-a*m^2-a*d*d^2-2a*m*d+b*m+b*d-c=0---------------------(ii) Again, Putting the value of m in polynomial, we have m^3-a*m^2+b*m-c=0-----------------------------------------(iii) Adding equation (i) & (ii), we have=>2*m^3+6*m*d^2-2a*m^2-2a*d^2+2b*m-2*c=0 =>2*(3^3-a*m^2+b*m-c)+6m*d^2-2*a*d^2=0 =>6*m*d^2=2*a*d^2 =>m=a/3------------------------------(iv) Now, Putting the value of m in the polynomial, =>(a/3)^3-a*(a/3)^2+b*a/3-c=0 =>a^3/27-a^3/9+a*b/3-c=0 =>a^3-3*a^3+9*a*b-27*c=0 =>2*a^3-9*a*b+27*c=0 (Proved) read less

As we know, sum of roots = -(coeff. of x2)/(coeff. of x3) = a Since the roots ( or zeros) of polynomial are in AP, then we can assume the roots to be A-D, A, A+D ( with D as common diff.) sum of roots = A-D+A+A+D = 3A =a therefore A = a/3 Since, A is a zero of the polynomial, then it should satisfy the eqn. put A=a/3 in eqn (a^3)/27 - (a^3)/9 + (a*b)/3 - c = 0 Hence, 2a3-9ab+27c = 0 read less