Let $$ f: [1 , \infty ) \rightarrow R$$ be a differentiable function. If $$6 \int_{1}^{x} f(t)dt=3x f(x)+ x^{3}-4$$ for all $$x\geq 1$$ then the value of $$f(2)-f(3)$$ is

Differentiate both sides of the given equation with respect to $$x$$. The left side is $$6\int_{1}^{x}f(t)\,dt$$. By the Fundamental Theorem of Calculus and the chain rule, its derivative is $$6f(x)$$. The right side is $$3x\,f(x)+x^{3}-4$$. Its derivative is $$3\bigl(f(x)+x\,f'(x)\bigr)+3x^{2}$$.

Therefore, we get $$6f(x)=3\bigl(f(x)+x\,f'(x)\bigr)+3x^{2}\quad-(1)$$

Expand and simplify $$(1)$$: $$6f(x)=3f(x)+3x\,f'(x)+3x^{2}$$ Subtract $$3f(x)$$ from both sides: $$3f(x)=3x\,f'(x)+3x^{2}$$ Divide by 3: $$f(x)=x\,f'(x)+x^{2}\quad-(2)$$

Rearrange $$(2)$$ into standard linear form: $$x\,f'(x)-f(x)=-x^{2}$$ Divide through by $$x$$ (for $$x\ge 1$$, $$x\neq 0$$): $$f'(x)-\frac{1}{x}\,f(x)=-x\quad-(3)$$

Identify the integrating factor for the linear equation $$(3)$$. We have $$\mu(x)=\exp\Bigl(-\int\frac{1}{x}\,dx\Bigr)=\exp(-\ln x)=\frac{1}{x}.$$ Multiply $$(3)$$ by $$\frac{1}{x}$$: $$\frac{1}{x}\,f'(x)-\frac{1}{x^{2}}\,f(x)=-1.$$ The left side is the derivative of $$\frac{f(x)}{x}$$. Hence, $$\frac{d}{dx}\Bigl(\frac{f(x)}{x}\Bigr)=-1.$$

Integrate both sides with respect to $$x$$: $$\frac{f(x)}{x}=-x+C$$ where $$C$$ is the constant of integration. Multiply by $$x$$: $$f(x)=-x^{2}+Cx\quad-(4)$$

To find $$C$$, use the condition at $$x=1$$ from the original equation: $$6\int_{1}^{1}f(t)\,dt=0=3\cdot1\cdot f(1)+1^{3}-4$$ gives $$3f(1)-3=0\implies f(1)=1.$$ Substitute into $$(4)$$: $$1=-1^{2}+C\cdot1=-1+C\implies C=2.$$

Thus the function is $$f(x)=-x^{2}+2x.$$ Now compute $$f(2)=-4+4=0,\quad f(3)=-9+6=-3.$$ Therefore, $$f(2)-f(3)=0-(-3)=3.$$

Final Answer: 3 (Option A)