A simple one

Go through the factorisation below, $\frac{a^{3}+b^{3}}{2}=4-3ab$ $a^{3}+b^{3}=8-6ab$ $a^{3}+b^{3}+(-2)^{3}=3ab(-2)$ $a^{3}+b^{3}+(-2)^{3}-3ab(-2)=0$ $(a+b-2)\cdot \frac{1}{2}[(a-b)^{2}+(b+2)^{2}+(2-a)^{2} ] =0$ We get that else $a+b-2=0$ or $(a-b)^{2}+(b+2)^{2}+(2-a)^{2} =0$

$\because a,b\in \mathbb{Z^{+}}$ only solutions we can obtain are,

$a+b=2 \Rightarrow (a,b)=(1,1)$

$(a-b)^{2}+(b+2)^{2}+(2-a)^{2}=0\Rightarrow$ No solutions for $a,b$ .

It yields that there is only one unique value for the answer.

$\therefore$ 0 suites for $a-b$ .

I am using the definition that $0$ is not a whole number ( more... ).

$\dfrac{a^3+b^3}{2} = 4 - 3ab \quad \Rightarrow a^3+b^3 = 8 - 6ab$ .

Since $a,b > 0\quad \Rightarrow a^3+b^3 = 8 - 6ab < 8\quad \Rightarrow a, b < 2 \quad \Rightarrow a=b=1 \\ \Rightarrow a - b = \boxed{0}$

This is a very easy question . Since it is Given that a and b are positive integers this implies RHS of the equation must be greater than 0 . Thus it implies ab < (4/3) . i.e. a and b must be less than 2 . Thus they both are = 1 . Hence solution . Easiest question i Faced :)