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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

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?

yes

Okie dokie thank you!

welcome

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?

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...

right

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?

x-3=0?

yes

therefore x = 3?

now solve for x

right

so there's your answer

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

yup and welcome