Question

If $a,b,c,d$ are four distinct real numbers which are in an A.P., then the smallest positive value of $k$ satisfying $2(a-b)+k(b-c{)}^2+(c-a{)}^3=2(a-d)+(b-d{)}^2+(c-d{)}^3$ is

Answer 1

Given : $a,b,c,d$ are in an A.P.
Assuming the common difference be $D$,
Now,
$2(a-b)+k(b-c{)}^2+(c-a{)}^3=2(a-d)+(b-d{)}^2+(c-d{)}^3\Rightarrow 2(-D)+k(-D{)}^2+(2D{)}^3=2(-3D)+(-2D{)}^2+(-D{)}^3\Rightarrow 4D+(k-4)D^2+9D^3=0\Rightarrow D(9D^2+(k-4)D+4)=0$
As $a,b,c,d$ are distinct, so $D\ne 0$
$\Rightarrow 9D^2+(k-4)D+4=0$
As $D\in R$
$\Rightarrow (k-4{)}^2-4(4\left)\right(9)\ge 0\Rightarrow k^2-8k-128\ge 0\Rightarrow (k-16\left)\right(k+8)\ge 0\Rightarrow k\in (-\infty ,-8]\cup [16,\infty )$

Hence, the smallest positive value of $k$ is $16$.