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add -8 to both sides
well i got to this part x^3-19=0
no \[ \large (x-3)^3=-8 \]

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Other answers:

k then
now take cubic roots of each side
x-3=2i?
yes
no \[ \large \sqrt[3]{(x-3)^3}=\sqrt[3]{-8}=-2 \]
then u have \[ \large x-3=-2 \]
ook so now why did you subtract 8 from both side instead of doing the parenthesis
if u had done that, in the end u would have had to use syntthetic division or something else to solve the equation. it is not a bad idea, but it is harder
so how do know when to do your method
how would i know when do apply your method
did they teach the order of operations?
nop
come on... for real??
oo my bad didnt read read the question right
well. take a look at this \[ \large (7-3)^8+8= \]
tell me step by step how would u compute this
if i were do do this i would find the difference of the number in parenthesis, then raise the result to the 8th power and then add it to 8
great
to solve an equation u do that but in reverse order: 1st undo +8 2nd undo ^3 3rd undo -3 got it?
hmm
did they (your teachers) taught you this?
i dont remember something like this lol

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