If $x$ is a positive real number, then the greatest value of $(7-x){(x+5)}^2,$ is 

Let $u=7-x$ and $v=x+5.$ Then,

$u+v=12$(constant)

Thus, we have to find the greatest value of $P=u{v}^2$ when $u-v=12$

Clearly, $P$ is the product of three factors. So, we express $u-v=12$ as the sum of three factors as given below. 

$u+2\left(\frac{v}{2}\right)=12$ or, $u+\frac{v}{2}+\frac{v}{2}=12$

$P^{\mathrm{\prime }}=u\left(\frac{v}{2}\right)\left(\frac{v}{2}\right)$ is greatest when $u=\frac{v}{2}$

Putting $u=\frac{v}{2}$ in $u+v=12$, we get 

$\begin{array}{l}\frac{u}{2}+v=12\Rightarrow v=4\\ u=\frac{v}{2}=4\end{array}$

Thus, $P\text{'}=u\left(\frac{v}{2}\right)\left(\frac{v}{2}\right)$ is greatest when $u=4$ and $v=8.$

Therefore, $P$ is greatest when $u=4$ and $v=8.$

Hence, $(7-x)(x+5{)}^2$ is greatest when $7-x=4$ and $x+5=8$

when $x=3.$

Greatest value$=(7-3)(3+5{)}^2=256$