Given: $f(x) = \log(3x + 3 - 2x - 2)$
We need to prove that: $\frac{(a+b)^2}{f(a)+f(b)} = -4$
First, we need to simplify $f(x)$:
$f(x) = \log(x+1)$ (by simplifying the expression inside the log)
Now, we can rewrite the expression we need to prove as:
$\frac{(a+b)^2}{\log(a+1) + \log(b+1)} = -4$
Using the property of logarithms, we can rewrite this as:
$\log((a+1)(b+1)) = \log(e^{-4})$
Simplifying further:
$(a+1)(b+1) = e^{-4}$
Expanding the left side and simplifying:
$ab + a + b + 1 = e^{-4}$
Now, we can use the identity $(a+b)^2 = a^2 + 2ab + b^2$:
$(a+b)^2 = a^2 + 2ab + b^2$
Dividing both sides by $f(a)+f(b)$, which is $\log(a+1) + \log(b+1)$:
$\frac{(a+b)^2}{\log(a+1) + \log(b+1)} = \frac{a^2 + 2ab + b^2}{\log(a+1) + \log(b+1)}$
Using the property of logarithms again, we can simplify the denominator:
$\frac{(a+b)^2}{\log((a+1)(b+1))} = \frac{a^2 + 2ab + b^2}{\log((a+1)(b+1))}$
Substituting the value we derived earlier for $(a+1)(b+1)$:
$\frac{(a+b)^2}{\log(e^{-4})} = \frac{a^2 + 2ab + b^2}{\log(e^{-4})}$
Simplifying:
$(a+b)^2 = a^2 + 2ab + b^2$
$4ab = -4$
Dividing both sides by 4:
$ab = -1$
Therefore, we have shown that $\frac{(a+b)^2}{f(a)+f(b)} = -4$ if $f(x) = \log(3x+3-2x-2)$.