anonymous
  • anonymous
(x-3)^3+8=0
Mathematics
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anonymous
  • anonymous
(x-3)^3+8=0
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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helder_edwin
  • helder_edwin
add -8 to both sides
anonymous
  • anonymous
well i got to this part x^3-19=0
helder_edwin
  • helder_edwin
no \[ \large (x-3)^3=-8 \]

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anonymous
  • anonymous
k then
helder_edwin
  • helder_edwin
now take cubic roots of each side
anonymous
  • anonymous
x-3=2i?
anonymous
  • anonymous
yes
helder_edwin
  • helder_edwin
no \[ \large \sqrt[3]{(x-3)^3}=\sqrt[3]{-8}=-2 \]
helder_edwin
  • helder_edwin
then u have \[ \large x-3=-2 \]
anonymous
  • anonymous
ook so now why did you subtract 8 from both side instead of doing the parenthesis
helder_edwin
  • helder_edwin
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
anonymous
  • anonymous
so how do know when to do your method
anonymous
  • anonymous
how would i know when do apply your method
helder_edwin
  • helder_edwin
did they teach the order of operations?
anonymous
  • anonymous
nop
helder_edwin
  • helder_edwin
come on... for real??
anonymous
  • anonymous
oo my bad didnt read read the question right
helder_edwin
  • helder_edwin
well. take a look at this \[ \large (7-3)^8+8= \]
helder_edwin
  • helder_edwin
tell me step by step how would u compute this
anonymous
  • anonymous
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
helder_edwin
  • helder_edwin
great
helder_edwin
  • helder_edwin
to solve an equation u do that but in reverse order: 1st undo +8 2nd undo ^3 3rd undo -3 got it?
anonymous
  • anonymous
hmm
helder_edwin
  • helder_edwin
did they (your teachers) taught you this?
anonymous
  • anonymous
i dont remember something like this lol

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