Question 798861
{{{ab^2=a*b*b}}} and {{{a^2b^2=a*a*b*b=a*b*a*b=(a*b)*(a*b)=(ab)*(ab)=(ab)^2}}}
There is no way ab^2={{{ab^2}}} could be interpreted as {{{a*a*b*b}}}.
Without brackets, the ^2 applies only to the variable or number right before it. That would include more than one character only if those characters are forming a single variable or a single number such as the square of the length of segment AB written as AB^2, or the square of 13 written as 13^2=169.


I like to use brackets when there could be any doubt.
Some could interpret -4ab*1/2ba^2 as {{{-4ab}}}{{{1/("2 b" a^2)=-4ab*1/2ba^2}}}.
If I meant that, I would write it as -4ab*1/(2ba^2).
 
 
I would interpret -4ab*1/2ba^2 as
-4 times a, times b, times 1, divided by 2, times b, and times {{{a^2}}}
{{{-4ab*1}}} ÷ {{2*ba^2=-4ab*1/2}}}{{{ba^2}}}
The same could be written as
{{{-4ab}}}{{{1/2}}}{{{ba^2=-4ab(1/2)ba^2=-4ab*0.5ba^2=-4*0.5*a*a^2*b*b=-2a^3b^2}}}


I would interpret -4ab*1/2ab^2 as
-4 times a, times b, times 1, divided by 2, times b, and times {{{a^2}}}
{{{-4ab*1}}}÷{{{2*ab^2=-4ab*1/2}}}{{{ab^2}}}
The same could be written as
{{{-4ab}}}{{{1/2}}}{{{ab^2 =-4ab(1/2)ab^2=-4ab*0.5ab^2=-4*0.5*a*a*b*b^2=-2a^2b^3}}}