Question

If the zeros of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ − 3abc + a²d = 0.

Answer 1

Let a - d, a and a + d be the zeros of the polynomial f(x). Then,

Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`

`a-d+a+a+d=(-3b)/a`

`3a=(-3b)/a`

`a=(-3b)/axx1/3`

`a=(-b)/a`

Since a is a zero of the polynomial f(x).

Therefore,

f(x) = ax³ + 3bx² + 3cx + d

f(a) = 0

f(a) = aa³ + 3ba² + 3ca + d

aa³ + 3ba² + 3ca + d = 0

`a((-b)/a)^3+3bxx((-b)/a)^2+3cxx((-b)/a)+d=0`

`axx(-b)/axx(-b)/axx(-b)/a+3xxbxx(-b)/axx(-b)/a+3xxcxx(-b)/a+d=0`

`(-b^3)/a^2+(3b^3)/a^2-(3cb)/a+d=0`

`(-b^3+3b^3-3abc+a^2d)/a^2=0`

2b³ − 3abc + a²d = 0 xa²

2b³ − 3abc + a²d = 0

Hence, it is proved that 2b³ − 3abc + a²d = 0