anonymous 4 years ago 3(x-3)^1/3 - 4(x-3)^4/3 + (x-3)^7/3 = 0 To make it easier, I substituted (x-3) with x, in which I got x^1/3(3 - 4x +x^2) ...but it can't be factored. Help me out! It's a fairly easy question I bet, but I need a nice way of solving it. Thank you! :)

1. anonymous

actually you might want to substitute x - 3 with a instead to avoid confusion

2. anonymous

but you're right anyway... except for the last part

3. anonymous

wait on second thought you're right

4. anonymous

and it can be factored... $\huge a^{\frac 13} (3 - 4a + a^2) \implies a^{\frac 13} (a^2 - 4a + 3)$ can you see it now?

5. anonymous

a^2 - 4a + 3 is factorable

6. anonymous

Oh okay! I kept thinking that it was a^2 + 4a - 3

7. anonymous

oh. no wonder

8. anonymous

yeah. that one isn't factorable

9. anonymous

so 3 - 4a + a^2 is a^2 - 4a + 3 ...right?

10. anonymous

yes

11. anonymous

Okie dokie thank you!

12. anonymous

welcome

13. anonymous

So wait, how would I figure out the final answer, I am at (x-3)^1/3 (x-7)(x-7)=0

14. anonymous

they shouldn't be equal...

15. anonymous

how did you factor a^2 - 4a + 3?

16. anonymous

I skipped a few steps. Remember a = (x-3) a^1/3 (a-1)(a-3) = 0 (x-3)^1/3 [(x-3-1)(x-3-3)] = 0 (x-3)^1/3 (x-4)(x-6) = 0

17. anonymous

sorry about my previous comment tho, I totally typed it wrong, my computer is lagging so bad.

18. anonymous

right. that's better

19. anonymous

So how do I solve the end?

20. anonymous

(x-3)^1/3 (x-4)(x-6) = 0

21. anonymous

that s the end

22. anonymous

isn't it?

23. anonymous

or do you want x?

24. anonymous

don't I make each = to zero

25. anonymous

Like, 2 of my answers will be 4 & 6...

26. anonymous

right

27. anonymous

now..just equate (x-3)^1/3 to 0

28. anonymous

I know but how :(

29. anonymous

$(x - 3)^{\frac 13} = 0$

30. anonymous

take the cube root of both sides..what happens?

31. anonymous

x-3=0?

32. anonymous

yes

33. anonymous

therefore x = 3?

34. anonymous

now solve for x

35. anonymous

right

36. anonymous

37. anonymous

Okay, so answer is 3,4,6 :) thanks!

38. anonymous

yup and welcome