Farhan's Algebraic Problem in finding values

The equation $a - b = 0$ simply means that $a = b$ . Substituting $a = b = x$ we find $2x^3 + 6x^2 = 8.$ It is easy to see that $x = 1$ is a solution. Divide out the factor $x-1$ : $0 = 2x^3 + 6x^2 - 8 = 2(x-1)(x^2 + 4x + 4) = 2(x-1)(x+2)^2,$ showing that $x = -2$ is the only other solution.

Therefore $ab + a + b + 1 = 1^2 + 1 + 1 + 1 = \boxed{4};\ \ \ \text{or}\ \ \ ab + a + b + 1 = (-2)^2 + 2\cdot (-2) + 1 = \boxed{1}.$

$a^3+6ab+b^3=8$ $a^3+b^3+3ab\times 2=2^3$ Compare this with $a^3+b^3+3ab(a+b)=(a+b)^3$ .

We get $a+b=2$ and $a-b=0$ . So $a=b=1$ , $ab+a+b+1=(a+1)(b+1)=2\times 2=4$ .