How do you simplify ${({e^3})^2} \times {e^2}$?

Hint:solve the brackets first and then solve the like terms to get the answer The given equation can be solved by using the BODMAS rule and hence we will be solving the brackets first and then the multiplication. To solve the brackets we will use the property ${({x^y})^z} = {x^{y \times z}}$. After which we will be left with two like terms with different powers. There we will use this property ${x^y} \times {x^z} = {x^{y + z}}$ to find a single term which will be our most simplified form of the given equation.

Complete step by step solution:
Here, the given equation is ${({e^3})^2} \times {e^2}$
using the property that the form ${({x^y})^z}$can be re written as ${x^{y \times z}}$on the given form we will get
\[ {({e^3})^2} \times {e^2} = {e^{3 \times 2}} \times {e^2} \\
\Rightarrow {e^6} \times {e^2} \\
\]
Now using another property which states that \[{x^y} \times {x^z}\]is equal to \[{x^{y + z}}\]on the above step will give us the most simplified answer
\[
{e^6} \times {e^2} = {e^{6 + 2}} \\
\Rightarrow {e^8} \\
\]
Therefore, the most simplified form of ${({e^3})^2} \times {e^2}$is \[{e^8}\].
Alternate method:
Here, the given equation is ${({e^3})^2} \times {e^2}$
which can also be written as ${({e^2} \times e)^2} \times {e^2}$
considering ${e^2}$=k,
$
{({e^2} \times e)^2} \times {e^2} = {(k \times \sqrt k )^2} \times k \\
\Rightarrow {k^2} \times k \times k = {k^4} \\
$
But ${e^2} = k\,$ therefore, putting the value of k in ${k^4}$we will get
${k^4} = {({e^2})^4}$
So, using the property that the form ${({x^y})^z}$can be re written as ${x^{y \times z}}$on the given form we will get
${({e^2})^4} = {e^{2 \times 4}} = {e^8}$
Therefore, the most simplified form of ${({e^3})^2} \times {e^2}$is \[{e^8}\].

Note: Be careful when you use both the properties because when the form is of the type ${({x^y})^z}$ we will have to multiply the subscripts whereas when the form is of \[{x^y} \times {x^z}\]we will have to add the subscripts and not the vice versa as the opposite will get you wrong every time.