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