$$\frac{(4.53-3.07)^{2}}{(3.07-2.15)(2.15-4.53)}+\frac{(3.07-2.15)^{2}}{(2.15-4.53)(4.53-3.07)}+\frac{(2.15-4.53)^{2}}{(4.53-3.07)(3.07-2.15)}$$ is simplified to

Let, a=3.07-2.15

b=2.15-4.53

c=4.53-3.07

So, a+b+c=3.07-2.15+2.15-4.53+4.53-3.07=0

So, $$\dfrac{(4.53-3.07)^{2}}{(3.07-2.15)(2.15-4.53)}+\dfrac{(3.07-2.15)^{2}}{(2.15-4.53)(4.53-3.07)}+\dfrac{(2.15-4.53)^{2}}{(4.53-3.07)(3.07-2.15)}$$ can be written as

=$$\dfrac{c^2}{ab}+\dfrac{a^2}{bc}+\dfrac{b^2}{ac}$$

=$$\dfrac{a^3+b^3+c^3}{abc}$$

Now, since, $$a+b+c=0$$

So, $$a^3+b^3+c^3=3abc$$ (By property)

So, the value of expression

=$$\dfrac{a^3+b^3+c^3}{abc}$$

=$$\dfrac{3abc}{abc}=3$$