Question 695456
{{{f(1) = 1^3+ a(1)^2 + b(1) -6 = a + b -5 }}}
f(1) = 0 therefore a + b = 5
{{{f(2) = 2^3+ a(2^2) + b(2) -6 = 4a + 2b + 2 }}}
f(2) = 20 therefore 4a + 2b + 2 = 20, transpose 2 then divide by -2
The result is -2a - b = -9, add to a + b = 5
The result is -a = -4, therefore a = 4 and b = 1. 

Answer: a = 4 and b = 1