## anonymous 4 years ago How do you solve 8x^3-27?

1. precal

it needs to be equal to something for you to solve it

2. anonymous

its equal to zero, sorry

3. anonymous

convert it to the form (a^3 + b^3) = (a+b)(a^2-ab + b^2)

4. anonymous

(2x + 3)(4x^2 - 6x + 9)...just distribute to get the final answer

5. anonymous

so do i set the first equation to zero and solve then use the quadratic formula to solve the second equation to find the roots?

6. anonymous

no! distribute....

7. anonymous

i'm lost

8. anonymous

oh wait...i was factoring :))) I should've been solving. Sorry for confusing you. Transpose 27 to the other side. 8x^3 = 27 get the cube root 2x = 3 x = 3/2

9. anonymous

there should be two more roots..

10. anonymous

that's okay, thank-you

11. anonymous

@cinar...i think this is a repetitive root

12. anonymous

no, they are complex

13. anonymous

$\frac34(-1+i \sqrt{3})\quad and \quad \frac34(-1-i \sqrt{3})\quad are \,also\, roots$