2023 IOQM/Problem 12

P(x) = x³ + ax² + bx + c.

a,b,c ∈ ℤ

And, c is odd.

p₁ = P(1) = (1)³ + a(1)² + b(1) + c p₁ = 1 + a + b + c

p₂ = P(2) = (2)³ + a(2)² + b(2) + c p₂ = 8 + 4a + 2b + c

p₃ = P(3) = (3)³ + a(3)² + b(3) + c p₃ = 27 + 9a + 3b + c

And, p₀ = P(0) = (0)³ + a(0)² + b(0) + c p₀ = c

Now, it is given that, p₁³ + p₂³ + p₃³ = 3p₁p₂p₃

this is a well known identity. If the SUM OF CUBES of 3 numbers is equal to THREE TIMES their product, then either their sum is 0 or the given numbers are EQUAL to each other. –> p₁ = p₂ = p₃

--> p₁ = p₂

--> 1 + a + b + c = 8 + 4a + 2b + c

==> 7 + 3a + b = 0 --------------> (1)

and, p₂ = p₃

--> 8 + 4a + 2b + c = 27 + 9a + 3b + c

==> 19 + 5a + b = 0 --------------> (2)

Now, subtract Equation (1) from (2)

12 + 2a = 0

==> a = (-6)

Using [a = (-6)] in equation (1)

7 + 3(-6) + b = 0

=> b = 11

Finally, calculating the value of -> [p₂ + 2p₁ - 3p₀]

p₂ + 2p₁ - 3p₀ = 8 + 4a + 2b + c + 2[1 + a + b + c] - 3[c]

= 8 + 4a + 2b + c + 2 + 2a + 2b + 2c - 3c

Rearranging the terms, we get

p₂ + 2p₁ - 3p₀ = 8 + 2 + 4a + 2a + 2b + 2b + c + 2c - 3c

= 10 + 6a + 4b

Using values of (a) and (b) which were calculated above

p₂ + 2p₁ - 3p₀ = 10 + 6(-6) + 4(11)

Solving further, we finally get

p₂ + 2p₁ - 3p₀ = 18

~ Arpit