(x-3)^3+8=0

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- anonymous

(x-3)^3+8=0

- schrodinger

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- helder_edwin

add -8 to both sides

- anonymous

well i got to this part x^3-19=0

- helder_edwin

no
\[ \large (x-3)^3=-8 \]

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- anonymous

k then

- helder_edwin

now take cubic roots of each side

- anonymous

x-3=2i?

- anonymous

yes

- helder_edwin

no
\[ \large \sqrt[3]{(x-3)^3}=\sqrt[3]{-8}=-2 \]

- helder_edwin

then u have
\[ \large x-3=-2 \]

- anonymous

ook so now why did you subtract 8 from both side instead of doing the parenthesis

- 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

so how do know when to do your method

- anonymous

how would i know when do apply your method

- helder_edwin

did they teach the order of operations?

- anonymous

nop

- helder_edwin

come on... for real??

- anonymous

oo my bad didnt read read the question right

- helder_edwin

well. take a look at this
\[ \large (7-3)^8+8= \]

- helder_edwin

tell me step by step how would u compute this

- 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

great

- 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

hmm

- helder_edwin

did they (your teachers) taught you this?

- anonymous

i dont remember something like this lol

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