Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be a function satisfying the relation $$4f(3-x)+3f(x)=x^{2}$$ for any real $$x$$. Find the value of $$f(27)-f(25)$$ to the nearest integer. (Here $$\mathbb{R}$$ denotes the set of real numbers.)

The given functional equation is

$$4f(3-x)+3f(x)=x^{2}\qquad\text{for all }x\in\mathbb{R}.\tag{-1}$$

Step 1: Write the same equation for $$3-x$$ instead of $$x$$.

Replace $$x$$ by $$3-x$$ in $$(\-1)$$:

$$4f\!\bigl(3-(3-x)\bigr)+3f(3-x)=(3-x)^{2}$$

Simplifying,

$$4f(x)+3f(3-x)=(3-x)^{2}.\tag{-2}$$

Step 2: Solve the linear system formed by $$(\-1)$$ and $$(\-2)$$.

From $$(\-1)$$: $$3f(x)+4f(3-x)=x^{2}.\tag{-1}$$

Multiply $$(\-1)$$ by $$4$$ and $$(\-2)$$ by $$3$$ to eliminate $$f(x)$$:

$$12f(x)+16f(3-x)=4x^{2}\tag{-3}$$

$$12f(x)+9f(3-x)=3(3-x)^{2}\tag{-4}$$

Subtract $$(\-4)$$ from $$(\-3)$$:

$$7f(3-x)=4x^{2}-3(3-x)^{2}.$$

Since $$(3-x)^{2}=x^{2}-6x+9$$,

$$7f(3-x)=4x^{2}-3(x^{2}-6x+9)=x^{2}+18x-27.$$

Therefore

$$f(3-x)=\frac{x^{2}+18x-27}{7}.\tag{-5}$$

Step 3: Obtain an explicit formula for $$f(x)$$.

Put $$(\-5)$$ back into $$(\-1)$$:

$$3f(x)=x^{2}-4f(3-x)=x^{2}-4\cdot\frac{x^{2}+18x-27}{7}.$$

Thus

$$3f(x)=\frac{7x^{2}-4x^{2}-72x+108}{7}=\frac{3x^{2}-72x+108}{7},$$

$$f(x)=\frac{3x^{2}-72x+108}{21}=\frac{x^{2}-24x+36}{7}.\tag{-6}$$

Step 4: Evaluate $$f(27)$$ and $$f(25)$$.

Using $$(\-6)$$:

$$f(27)=\frac{27^{2}-24\cdot27+36}{7}=\frac{729-648+36}{7}=\frac{117}{7}=16.7142857\ldots$$

$$f(25)=\frac{25^{2}-24\cdot25+36}{7}=\frac{625-600+36}{7}=\frac{61}{7}=8.7142857\ldots$$

Step 5: Compute the required difference.

$$f(27)-f(25)=\frac{117}{7}-\frac{61}{7}=\frac{56}{7}=8.$$

To the nearest integer, the value is $$\mathbf{08}$$.