## mskyeg 2 years ago What are the real or imaginary solutions of the polynomial equation? x3 = 216

1. mskyeg

2. hartnn

$$x^3-216=0 \\ x^3-6^3=0 \\ \text{now apply }a^3-b^3 formula$$

3. mskyeg

how do i do that

4. mskyeg

like what does it mean

5. hartnn

know the formula ?

6. mskyeg

like what do i plug intohe formula

7. mskyeg

*into the

8. hartnn

a=x,b=6

9. mskyeg

how does that help all i get is x^3 - 216

10. hartnn

thats because u don't know that formula

11. mskyeg

oooh i thought that WAS the formula

12. hartnn

$$\large a^3-b^3=(a-b)(a^2+ab+b^2)$$

13. hartnn

now put a=x,b=6

14. hartnn

and then equate it to 0

15. hartnn

confused ?

16. hartnn

$$(x-6)(x^2+6x+36)=0 \\ (x-6)=0,(x^2+6x+36)=0 \\ x=6,\text{use quadratic formula for x^2+6x+36=0}$$

17. hartnn

Compare your quadratic equation with $$ax^2+bx+c=0$$ find a,b,c then the two roots of x are: $$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

18. mskyeg

I tried but i got a really weird answer

19. hartnn

what u got, a,b,c as ?

20. mskyeg

a=6 b=-3+i square root 54 c=-3-i square root 54

21. hartnn

???

22. mskyeg

oh a=1 b=6 and c=36

23. hartnn

yes

24. hartnn

so two other values of x are ?

25. hartnn

one obviously is x=6

26. mskyeg

thats what u have u work out using $$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$ u have a,b,c values correct.