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anonymous
 4 years ago
(x+3)^2  (3x1)^2 = x^3 + 4?
i've tried this three times and I still get the wrong answer!
anonymous
 4 years ago
(x+3)^2  (3x1)^2 = x^3 + 4? i've tried this three times and I still get the wrong answer!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x+3)^2  (3x  1)^2 = x^3 + 4\]

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1unless you know the cubic formula, your best bet is to use a graphing calculator to find the approximate solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we're not allowed to use a calculator. the answer is supposed to come out to be  2/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but I don't know how to get there

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1ok so there's a typo somewhere in the problem then, my guess is that x^3 is one of the typos

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait, I made a mistake typing out the problem. it's actually \[(x+3)^3\] as the first one

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1oh ok thx, one sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \[(x+3)^3  (3x  1)^2 = x^3 + 4\]

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large (x+3)^3 (3x1)^2 = x^3 + 4\] \[\large x^3+9x^2+27x+27 (9x^26x+1) = x^3 + 4\] \[\large x^3+9x^2+27x+27 9x^2+6x1 = x^3 + 4\] \[\large x^3+9x^2+27x+27 9x^2+6x1  x^3  4=0\] \[\large 33x+22=0\] \[\large 33x=22\] \[\large x=\frac{22}{33}\] \[\large x=\frac{2}{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you! i see exactly where I went wrong.
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