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

We are given the functional equation

$$4f(3-x)+3f(x)=x^{2}\qquad\forall x\in\mathbb{R}$$ $$-(1)$$

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

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

Equations $$-(1)$$ and $$-(2)$$ form a linear system in the two unknowns $$f(x)$$ and $$f(3-x)$$:

$$\begin{cases} 3f(x)+4f(3-x)=x^{2} &\qquad -(1)\\[4pt] 4f(x)+3f(3-x)=(3-x)^{2} &\qquad -(2) \end{cases}$$

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

$$\begin{aligned} &(9f(x)+12f(3-x))-(16f(x)+12f(3-x))\\ &=3x^{2}-4(3-x)^{2} \end{aligned}$$

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

Compute the right‐hand side:

$$(3-x)^{2}=x^{2}-6x+9$$
$$4(3-x)^{2}=4x^{2}-24x+36$$
$$3x^{2}-4(3-x)^{2}=3x^{2}-(4x^{2}-24x+36)=x^{2}-24x+36$$

Hence

$$7f(x)=x^{2}-24x+36\quad\Longrightarrow\quad f(x)=\frac{x^{2}-24x+36}{7}$$ $$-(3)$$

Now evaluate at the required points.

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

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

Therefore

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

The value rounded to the nearest integer is $$8$$.

Answer: 08