Question

if the ratios of the polynomial ax3 +3bx2+3cx + d are in A.P, prove that 2b3 - 3abc +a2d= 0

Answer 1

Take a look at the pic.

Answer 2

solution :-


f(x) = ax³ + 3bx² + 3cx + d

Let the zeroes of the polynomial be (l – m), l and l + m.

∴ f(x) = ax³ + 3bx² + 3cx + d

⇒ [ x – (l – m) ] [ (x – l) (x – (l + m)) ] = ax³ + 3bx² + 3cx + d

⇒ (x – l) [ (x – (l – m)) ] [ x – (l + m) ]= ax³+ 3bx² + 3cx + d

⇒ (x – l) [ (x² – x (l + m) – x (l – m) – (l² –m²) ]= ax³ + 3bx² + 3cx + d

⇒ (x – l) [ (x² – lx – mx – lx + mx – (l² – m²) ] = ax³ + 3bx² + 3cx + d

⇒ (x – l) [ (x² – 2lx – (l² –m²) ] = ax³+ 3bx² + 3cx + d

⇒ x³ – 2lx²– x (l² –m²) – lx² + 2l2x + l (l² – m²) = ax³ + 3bx² + 3cx + d

⇒ x³ – 3lx² – x (l² –m²) + l (l² – m²) = ax³ + 3bx² + 3cx + d

comparing the coefficients of like powers on both sides of the given equation, we get
a = 1 [ on comparing the coefficient of x3]

so, 3b = – 3l [ on comparing the coefficient of x²]

⇒ b = –l
⇒ l = –b

so, 3c = l² + m² [ on comparing the coefficient of x]

=> c = l²+ m²/3

and d = l (l² – m²) [ on comparing the coefficient of constant term]

Now,
To Prove: 2b³– 3abc + a2d = 0

L.H.S. = 2(– l)³– 3(1) (– l)


( l² +m²/ 3)+(1)2 · l (l² – m²)

= –2beta + l beta + m²l + beta – m²l

= 0 = R.H.S. [ Hence proved ]