Question 3
c = (a + bi)³ - 107i
c + 107i = (a + bi)³
c + 107i = a³ + 3a²(bi) + 3a(bi)² + (bi)³
(Based on the result (x + y)³ = x³ + 3x²y + 3xy² + y³ . You can use binomial theorem or expand it manually to verify it)
c + 107i = a³ + 3a²bi + 3ab²i² + b³i³
c + 107i = a³ + 3a²bi + 3ab² (-1) + b³i(i²)
c + 107i = a³ + 3a²bi - 3ab² - b³i
c + 107i = a³ - 3ab² + 3a²bi - b³i
c + 107i = a³ - 3ab² + b(3a² - b²)i

Comparing real parts on both sides,
c = a³ - 3ab²
Comparing imaginary parts on both sides,
107 = b(3a² - b²)
Now 107 is prime. 107 = 107 × 1, a product of 2 positive integers.
Since 107 = b(3a² - b²) and b is a positive integer, b has to be 1 or 107. We can't factorise 107 into any other two positive integer factors.
① If b = 107, then 3a² - b² = 1
3a² = 1 + b² = 1 + 107² = 1 + 11449 = 11450
Then a² = 11450 ÷ 3 = 3816⅔
Recall that a is a positive integer. So its square (a²) has to be a positive integer as well. So a² cannot equal 3816⅔. We reject this option.

② b = 1, then 3a² - b² = 107
3a² - 1² = 107
3a² = 108
a² = 36
a = √36 = 6 (we don't consider -6 since a is positive)

Therefore, c = a³ - 3ab² = 6³ - 3(6)(1²)
= 216 - 18
= 198