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dinainjune

  • 3 years ago

how to solve this a^3>-1 and the answer is a<-1 , can anyone explain to me?

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  1. saifoo.khan
    • 3 years ago
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    Yea!!

  2. Ishaan94
    • 3 years ago
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    a^3 > -1 a > -1

  3. hahd
    • 3 years ago
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    cube toot of a is?

  4. Nick_Black
    • 3 years ago
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    a^3>-1 a>-1

  5. Ishaan94
    • 3 years ago
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    toot !? lol

  6. saifoo.khan
    • 3 years ago
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    take cube root on both sides.

  7. saifoo.khan
    • 3 years ago
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    the cube root of -1 is -1.

  8. hahd
    • 3 years ago
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    ull end up with -1 again hence the answer

  9. Nick_Black
    • 3 years ago
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    no saif its -1

  10. saifoo.khan
    • 3 years ago
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    what nick?

  11. Ishaan94
    • 3 years ago
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    saif said the same

  12. Nick_Black
    • 3 years ago
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    -1^3=-1

  13. saifoo.khan
    • 3 years ago
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    Yea!

  14. dinainjune
    • 3 years ago
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    yes, it's cube root. and i still confuse

  15. Ishaan94
    • 3 years ago
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    or there is another way too a^3 > -1 a^3 +1 > 0 (a + 1)(a^2 + 1 -a)> 0 ^ ^ real complex a >-1

  16. jim_thompson5910
    • 3 years ago
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    a^3 > -1 a^3 + 1 > 0 (a+1)(a^2-a+1) > 0 .. factor with the sum of cubes formula Since a^2-a+1 is ALWAYS positive (look at the graph of a^2-a+1 to see this or complete the square on a^2-a+1), this means that a^2-a+1 has NO influence on the sign of the entire expression. So everything is dependent on the factor a+1 So because the entire expression is positive, and the sign depends on a+1, we know for sure that a+1 > 0 a+1 > 0 a > -1 So the solution is a > -1

  17. Ishaan94
    • 3 years ago
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    but I am sure you don't want the complex thing so just so a>-1

  18. jim_thompson5910
    • 3 years ago
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    make sure that there are no typos

  19. Ishaan94
    • 3 years ago
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    Good Job Jim

  20. dinainjune
    • 3 years ago
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    okay, thank you all :)

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