Question

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! :)

Answer 1

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

Answer 2

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

Answer 3

wait on second thought you're right

Answer 4

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?

Answer 5

a^2 - 4a + 3 is factorable

Answer 6

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

Answer 7

oh. no wonder

Answer 8

yeah. that one isn't factorable

Answer 9

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

Answer 10

yes

Answer 11

Okie dokie thank you!

Answer 12

welcome

Answer 13

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

Answer 14

they shouldn't be equal...

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

right. that's better

Answer 19

So how do I solve the end?

Answer 20

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

Answer 21

that
s the end

Answer 22

isn't it?

Answer 23

or do you want x?

Answer 24

don't I make each = to zero

Answer 25

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

Answer 26

right

Answer 27

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

Answer 28

I know but how :(

Answer 29

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

Answer 30

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

Answer 31

x-3=0?

Answer 32

yes

Answer 33

therefore x = 3?

Answer 34

now solve for x

Answer 35

right

Answer 36

so there's your answer

Answer 37

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

Answer 38

yup and welcome