Let (3-x) = a & (2-x) = b

Then the equation becomes,

a⁴ + b⁴ = (a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

2ab(2a² + 3ab + 2b²) = 0

Substituting the values of a & b, we get

x = 2,3 or 7x² - 35x +44 = 0

D = 1225 - 1232 = -7 < 0

Hence, (c) 2 imaginary & 2 real roots.