(ab^2c)^-1 = a^-1b^-2c^-1
a^8b^2c^-4b^3 = a^8b^5c^-4
a^-1b^-2c^-1 / a^8b^5c^-4 = a^-9b^-7c^3
= c^3 / a^9b^7
ok i get it now.
for this question..
3square root of 1/2x+1 +2=5
the +2=5 is out of the square root.
so do i get rid of the square root first or subtract 2 from 5
do the reverse of PEMDAS
3√1/(2x+1) + 2 = 5
3/√(2x+1) = -3
1/√(2x+1) = -1
1 = -√(2x+1)
1 = 2x+1
0 = 2x
x = 0
3/√1 + 2 = 5
Rather a poor exercise, imho
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