1. anonymous

$P(x)= x^{6}-729$

2. welshfella

x^6 - 729 = (x^3)^2 - 27^2 = (x^3 - 27)(x^3 + 27)

3. welshfella

now you can factor each parentheses using the cube identities a^3 + b^3 = (a + b)(a^2 - ab + b^2) and a^3 - b^3 = (a - b)(a^2 + ab + b^2)

4. welshfella

putting a^3 = x^3 and b^3 = 27 where a= x and b = 3

5. welshfella

to find the zeros let (x^3 - 27)( x^3 + 27) = 0 and solve for x

6. anonymous

$(x+3)\times (x^{2}-3x+9)\times (x-3)\times (x ^{2}+3x+9)$

7. anonymous

is that right?.. for the zeros.. should i just use the the quadratic equations to get the remained zeros?

8. welshfella

yes that's it

9. anonymous

AAAHH thank you!

10. welshfella

hmm maybe its better to find the zeroes using the last expression x = 3 and -3 are obvious roots to find the other 4 roots you need to solve x^2 - 3x + 9 = 0 and x^2 + 3x + 9 = 0

11. anonymous

could i solve them with the quadratic equation?

12. welshfella

they will all be complex roots

13. welshfella

14. anonymous

cool! thank you.. you were really helpful!

15. welshfella

yw

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