## a123 4 years ago (x+3)^2 - (3x-1)^2 = x^3 + 4? i've tried this three times and I still get the wrong answer!

1. a123

$(x+3)^2 - (3x - 1)^2 = x^3 + 4$

2. a123

just to clarify

3. jim_thompson5910

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

4. a123

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

5. a123

but I don't know how to get there

6. jim_thompson5910

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

7. a123

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

8. jim_thompson5910

oh ok thx, one sec

9. a123

so $(x+3)^3 - (3x - 1)^2 = x^3 + 4$

10. jim_thompson5910

$\large (x+3)^3- (3x-1)^2 = x^3 + 4$ $\large x^3+9x^2+27x+27- (9x^2-6x+1) = x^3 + 4$ $\large x^3+9x^2+27x+27- 9x^2+6x-1 = x^3 + 4$ $\large x^3+9x^2+27x+27- 9x^2+6x-1 - x^3 - 4=0$ $\large 33x+22=0$ $\large 33x=-22$ $\large x=-\frac{22}{33}$ $\large x=-\frac{2}{3}$

11. a123

thank you! i see exactly where I went wrong.