## magepker728 (x-3)^3+8=0 one year ago one year ago

1. helder_edwin

add -8 to both sides

2. magepker728

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

3. helder_edwin

no $\large (x-3)^3=-8$

4. magepker728

k then

5. helder_edwin

now take cubic roots of each side

6. magepker728

x-3=2i?

7. KE8717504

yes

8. helder_edwin

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

9. helder_edwin

then u have $\large x-3=-2$

10. magepker728

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

11. 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

12. magepker728

so how do know when to do your method

13. magepker728

how would i know when do apply your method

14. helder_edwin

did they teach the order of operations?

15. magepker728

nop

16. helder_edwin

come on... for real??

17. magepker728

18. helder_edwin

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

19. helder_edwin

tell me step by step how would u compute this

20. magepker728

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

21. helder_edwin

great

22. helder_edwin

to solve an equation u do that but in reverse order: 1st undo +8 2nd undo ^3 3rd undo -3 got it?

23. magepker728

hmm

24. helder_edwin

did they (your teachers) taught you this?

25. magepker728

i dont remember something like this lol