A algebra problem by Norwyn Kah

squaring $a+b+c=2$ will become $a^2+b^2+c^2+2ab+2ac+2bc=4$

comparing this equation with $a^2+b^2+c^2=2$ will give us the equation $ab+bc+ac=1$

expanding the equation we need to find will become $\frac{a^3+b^3+c^3-2a^2-2b^2-2c^2+a+b+c}{abc}$

$\Rightarrow\frac{a^3+b^3+c^3-2(2)+2}{abc}$

$\Rightarrow\frac{a^3+b^3+c^3-2}{abc}$

using the identity $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)$

$\Rightarrow a^3+b^3+c^3=8-3(2-c)(2-b)(2-a)$

$\Rightarrow a^3+b^3+c^3=8-3(8-4(a+b+c)+2(ab+bc+ac)-abc)$

$\Rightarrow a^3+b^3+c^3=8-3(8-4(2)+2(1)-abc)$

$\Rightarrow a^3+b^3+c^3=8-3(2-abc)$

$\Rightarrow a^3+b^3+c^3=2+3abc$

substituting the value of $a^3+b^3+c^3$ in $\frac{a^3+b^3+c^3-2}{abc}$

$\Rightarrow\frac{2+3abc-2}{abc}$

$\Rightarrow\frac{3abc}{abc}$

$\Rightarrow\boxed{3}$