#color(blue)("Note on method: ")#
#color(brown)("As you become more practised at this you will start to use")##color(brown)("shortcuts. This greatly reduces the amount of work and is much faster.")#
Suppose we have the variable #x# and it is written as #x^-2#.
Now write it as #x^(-2)/1#. Although this is not normally done it is correct.
Being raised to the power of (-2) basically means that it is the same as #1/(x^(2))#. Notice that after being moved the -2 becomes +2.
Suppose we have the variable written as #1/(x^(-2))#. This is the same as #(x^(2))/1#
#color(brown)("The conclusion is that if the index is negative then the value it is")#
#color(brown)("applied to may be moved to the other side of the line.")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Applying this to your question")#
The #2b^2# is within brackets. The negative index is applied to the brackets so this means it is applied to all of what is inside those brackets, i.e. #2b^2#
Given: #color(brown)(b^8(2b^2)^(-2))#
Write as: #b^8/((2b^2)^2)#
Note that because we are 'working on' the brackets as a whole we do not change its contents. That is: #2b^2" remains " 2b^2#.
Write as: #b^8/(2b^2xx2b^2)#
#=b^8/(4xxb^2xxb^2)#
# = b^8/(4b^((2+2)))#
#=b^8/(4b^4)#
But #b^8=b^4xxb^4# so by substitution we have:
#=(b^4xxb^4)/(4b^4)#
#=b^4/b^4xxb^4/4#
But #b^4/b^4 = 1# so we now have:
#1xxb^4/4 = b^4/4#
