Question

(x+3)^2 - (3x-1)^2 = x^3 + 4?
i've tried this three times and I still get the wrong answer!

Answer 1

\[(x+3)^2 - (3x - 1)^2 = x^3 + 4\]

Answer 2

just to clarify

Answer 3

unless you know the cubic formula, your best bet is to use a graphing calculator to find the approximate solutions.

Answer 4

we're not allowed to use a calculator. the answer is supposed to come out to be - 2/3

Answer 5

but I don't know how to get there

Answer 6

ok so there's a typo somewhere in the problem then, my guess is that x^3 is one of the typos

Answer 7

wait, I made a mistake typing out the problem. it's actually \[(x+3)^3\] as the first one

Answer 8

oh ok thx, one sec

Answer 9

so \[(x+3)^3 - (3x - 1)^2 = x^3 + 4\]

Answer 10

\[\large (x+3)^3- (3x-1)^2 = x^3 + 4\]


\[\large x^3+9x^2+27x+27- (9x^2-6x+1) = x^3 + 4\]


\[\large x^3+9x^2+27x+27- 9x^2+6x-1 = x^3 + 4\]


\[\large x^3+9x^2+27x+27- 9x^2+6x-1 - x^3 - 4=0\]


\[\large 33x+22=0\]


\[\large 33x=-22\]


\[\large x=-\frac{22}{33}\]


\[\large x=-\frac{2}{3}\]

Answer 11

thank you! i see exactly where I went wrong.