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helder_edwin
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add -8 to both sides
magepker728
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well i got to this part x^3-19=0
helder_edwin
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no
\[ \large (x-3)^3=-8 \]
magepker728
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k then
helder_edwin
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now take cubic roots of each side
magepker728
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x-3=2i?
KE8717504
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yes
helder_edwin
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no
\[ \large \sqrt[3]{(x-3)^3}=\sqrt[3]{-8}=-2 \]
helder_edwin
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then u have
\[ \large x-3=-2 \]
magepker728
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ook so now why did you subtract 8 from both side instead of doing the parenthesis
helder_edwin
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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
magepker728
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so how do know when to do your method
magepker728
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how would i know when do apply your method
helder_edwin
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did they teach the order of operations?
magepker728
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nop
helder_edwin
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come on... for real??
magepker728
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oo my bad didnt read read the question right
helder_edwin
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well. take a look at this
\[ \large (7-3)^8+8= \]
helder_edwin
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tell me step by step how would u compute this
magepker728
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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
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great
helder_edwin
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to solve an equation u do that but in reverse order:
1st undo +8
2nd undo ^3
3rd undo -3
got it?
magepker728
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hmm
helder_edwin
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did they (your teachers) taught you this?
magepker728
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i dont remember something like this lol