Not long enough

$\small \left(a + \dfrac{1}{a}\right)^2 + \left(b + \dfrac{1}{b}\right)^2 + \left(ab + \dfrac{1}{ab} \right)^2 - \left(a + \dfrac{1}{a}\right)\left(b + \dfrac{1}{b}\right)\left(ab + \dfrac{1}{ab}\right) \\ = \small \left(a + \dfrac{1}{a}\right)^2 + \left(b + \dfrac{1}{b}\right)^2 + \left(ab + \dfrac{1}{ab} \right)^2 - \left(ab + \dfrac{a}{b} + \dfrac{b}{a} + \dfrac{1}{ab}\right)\left(ab + \dfrac{1}{ab}\right) \\ = \small \left(a + \dfrac{1}{a}\right)^2 + \left(b + \dfrac{1}{b}\right)^2 + \left(ab + \dfrac{1}{ab} \right)^2 - \left(ab + \dfrac{1}{ab} \right)^2 - \left(\dfrac{a}{b} + \dfrac{b}{a} \right)\left(ab + \dfrac{1}{ab}\right) \\ = \small \left(a + \dfrac{1}{a}\right)^2 + \left(b + \dfrac{1}{b}\right)^2 - \left(\dfrac{a}{b} + \dfrac{b}{a} \right)\left(ab + \dfrac{1}{ab}\right) \\ = \small a^2 + 2 + \dfrac{1}{a^2} + b^2 + 2 + \dfrac{1}{b^2} - \left(a^2 + \dfrac{1}{b^2} + b^2 + \dfrac{1}{a^2} \right) \\ = \boxed{4}$

Because the given equation is true for all non-zero real numbers , put a=1,b=1, you will get 4+4+4-8 =4

Let a and b be non zero numbers. Then work it out like this:

C = $(a + \frac{1}{a})^{2}$ = $a^{2} + 2 + \frac{1}{a^{2}}$

D = $(b + \frac{1}{b})^{2} = b^{2} + 2 + \frac{1}{b^{2}}$

E = $(ab + \frac{1}{ab})^{2} = a^{2}b^{2} + 2 + \frac{1}{a^{2}b^{2}}$

F = $(a + \frac{1}{a})(b + \frac{1}{b}) = ab + \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}$

G = $(ab + \frac{1}{ab})$

F * G = $a^{2}b^{2} + 1 + a^{2} + \frac{1}{b^{2}} + b^{2} + \frac{1}{a^{2}} + 1 + \frac{1}{a^{2}b^{2}}$

C + D + E - F * G = $a^{2}b^{2} + a^{2} + b^{2} + \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{a^{2}b^{2}} + 6 - a^{2}b^{2} - a^{2} - b^{2} - \frac{1}{a^{2}} - \frac{1}{b^{2}} - \frac{1}{a^{2}b^{2}} - 2$

= 6 - 2

= 4