## mazehaq 3 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. lgbasallote

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

2. lgbasallote

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

3. lgbasallote

wait on second thought you're right

4. lgbasallote

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. lgbasallote

a^2 - 4a + 3 is factorable

6. mazehaq

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

7. lgbasallote

oh. no wonder

8. lgbasallote

yeah. that one isn't factorable

9. mazehaq

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

10. lgbasallote

yes

11. mazehaq

Okie dokie thank you!

12. lgbasallote

welcome

13. mazehaq

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

14. lgbasallote

they shouldn't be equal...

15. lgbasallote

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

16. mazehaq

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. mazehaq

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

18. lgbasallote

right. that's better

19. mazehaq

So how do I solve the end?

20. mazehaq

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

21. lgbasallote

that s the end

22. lgbasallote

isn't it?

23. lgbasallote

or do you want x?

24. mazehaq

don't I make each = to zero

25. mazehaq

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

26. lgbasallote

right

27. lgbasallote

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

28. mazehaq

I know but how :(

29. lgbasallote

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

30. lgbasallote

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

31. mazehaq

x-3=0?

32. lgbasallote

yes

33. mazehaq

therefore x = 3?

34. lgbasallote

now solve for x

35. lgbasallote

right

36. lgbasallote

37. mazehaq

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

38. lgbasallote

yup and welcome