Question

a, b, c, d are four distinct real numbers and they are in A.P. If $2(a-b)+x(b-c{)}^2+(c-a{)}^3=2(a-d)+(b-d{)}^2+(c-d{)}^3$ then prove that $x\ge 16$ or $x\le -8$.

Answer 1

Let D be common difference of A.P. where D$\ne 0$ then $b-a=D,c-a=2D,c-d=-D,b-d=-2D$ etc.
Thus the given equation can be written as $-2D+x{D}^2+(2D{)}^3=2(-3D)+(-2D{)}^2+(-D{)}^3$
Since $D\ne 0$, we can cancel D and arranging as a quadratic in D, we have
$9D^2+(x-4)D+4=0$
Since Dis real $\therefore \Delta \ge 0$ or $(x-4{)}^2-144\ge 0$ or $(x-16)(x+8)\ge 0$
$[\because p^2-Q^2=(P+Q\left)\right(P-Q\left)\right]$
$\therefore x\le -8$ or $x\ge 16$.