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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! :)

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actually you might want to substitute x - 3 with a instead to avoid confusion
but you're right anyway... except for the last part
wait on second thought you're right

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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?
a^2 - 4a + 3 is factorable
Oh okay! I kept thinking that it was a^2 + 4a - 3
oh. no wonder
yeah. that one isn't factorable
so 3 - 4a + a^2 is a^2 - 4a + 3 ...right?
Okie dokie thank you!
So wait, how would I figure out the final answer, I am at (x-3)^1/3 (x-7)(x-7)=0
they shouldn't be equal...
how did you factor a^2 - 4a + 3?
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
sorry about my previous comment tho, I totally typed it wrong, my computer is lagging so bad.
right. that's better
So how do I solve the end?
(x-3)^1/3 (x-4)(x-6) = 0
that s the end
isn't it?
or do you want x?
don't I make each = to zero
Like, 2 of my answers will be 4 & 6...
now..just equate (x-3)^1/3 to 0
I know but how :(
\[(x - 3)^{\frac 13} = 0\]
take the cube root of both sides..what happens?
therefore x = 3?
now solve for x
so there's your answer
Okay, so answer is 3,4,6 :) thanks!
yup and welcome

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