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dddan

  • 2 years ago

how do you factor 27y^3+8?

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  1. hartnn
    • 2 years ago
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    \(\huge a^3+b^3=(a+b)(a^2-ab+b^2)\)

  2. mukushla
    • 2 years ago
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    and here let a=3y and b=2

  3. hartnn
    • 2 years ago
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    write 27y^3 as(3y)^3 and ^^

  4. dddan
    • 2 years ago
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    alright but why is a=3 and b=2?

  5. hartnn
    • 2 years ago
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    nopes, a=3y, not 3

  6. hartnn
    • 2 years ago
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    and u had 8, u can write that as 2^3

  7. hartnn
    • 2 years ago
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    so 27y^3+8= (3y)^3+2^3 got this?

  8. dddan
    • 2 years ago
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    yea so u just plug it into the equation and get (3+2)(3^2-(3)(2)+2^2) right?....

  9. hartnn
    • 2 years ago
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    NOT 3+2 3y+2 a=3y

  10. dddan
    • 2 years ago
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    alright so then u just solve- so i got (3y+2)(9y^2-6y+4) i this correct?

  11. hartnn
    • 2 years ago
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    yes! that is correct :)

  12. dddan
    • 2 years ago
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    kool thanks!

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