Why differentiating a function is similar to differentiating its log function? $f(6;p)=\binom{25}{6}p^6(1-p)^6,\quad 0\le p\le1$ 
I have three questions:
$(1)$To find the relative maxima of p, the process is to take the derivative of the function with respect to $p$ equal to $0$ and solving the resulting equation for $p$. 

For easy computation we take the log of the function & differentiate the log function with respect to $p$ equal to $0$ and solving the resulting equation for $p$. 

$\bullet$"My Question Is Why $\frac{d}{dp}f(6;p)$ is similar to $\frac{d}{dp}\log f(6;p)$"
(2) After the step(1) i found that $p=\frac{6}{25}$ is the only root.

But on the book, which i am reading, is written that $p=0,1,\frac{6}{25}$  are the roots.

$\bullet$"How p=0,1 can be root? The denominator $p(1-p)$ becomes $0$ when i do cross multiplication."
(3)To check which value of $p$ maximize $f(6;p)$ , we take the second derivative of the function with respect to $p$ and set $p=0,1,\frac{6}{25}$ respectively. If the second derivative is less than $0$, then $p=$ that value maximize the function  .
for $p=\frac{6}{25}$ the second derivative is $-125.56<0$ , so $ p=\frac{6}{25} $ gives maximum.

i also got for $p=0,1$ the second derivative is $\quad-\infty$ , so $ p=0,1 $  also give maximum.
  But the book commented out that $p=0,1$ give a minimum.

$\bullet$"How  $p=0,1$ give a minimum?"

I agree with André Nicolas that $0$ and $1$ are probably meant to be regarded as endpoints, and therefore should not be treated with the second derivative test.  But even if you take the problem at face value, and consider $p$ to lie in the interval $(-\infty,\infty),$ which is what the book seems to do, there are issues at $p=0,1$ that are worth getting right.
The book's solution is suggesting logarithmic differentiation.  The idea is that
$$\frac{d}{dp}\log f(p)=\frac{f'(p)}{f(p)},$$
and therefore
$$f'(p)=f(p)\frac{d}{dp}\log f(p).$$
This is useful since, as you found, the derivative of $\log f(p)$ is much easier to compute than would be the derivative of $f(p)$ directly.  Nevertheless, you have to remember to multiply the derivative of $\log f(p)$ by $f(p)$ in order to get $f'(p)$.  When you do that, you find that $0$ and $1$ are indeed roots of $f'(p).$
To compute the second derivative, you can use logarithmic differentiation again, but you should make sure to simplify $f'(p)$ first.  That is, make sure to express $f'(p)$ as a product of factors before taking the logarithm again.
What you will find at the end of all this is that $f''(0)=f''(1)=0.$  When $f''(p)=0$ at a critical point of $f,$ it is not possible to draw any conclusion from the second derivative about whether that critical point is a relative minimum or maximum.  It is necessary to use the first derivative for this.  What you will find is that $p=0$ is a relative minimum and that $p=1$ is neither a relative minimum nor a relative maximum.  On the interval $[0,1],$ however, both $p=0$ and $p=1$ are absolute minima.  The graph should make this clear.
