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lokiando
 one year ago
Which product will result in a sum or difference of cubes?
a) (x + 7)(x^2 – 7x + 14)
b) (x + 8)(x^2 + 8x + 64)
c) (x – 9)(x^2 + 9x + 81)
d) (x – 10)(x^2 – 10x + 100)
lokiando
 one year ago
Which product will result in a sum or difference of cubes? a) (x + 7)(x^2 – 7x + 14) b) (x + 8)(x^2 + 8x + 64) c) (x – 9)(x^2 + 9x + 81) d) (x – 10)(x^2 – 10x + 100)

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4sum of cubes formula \[(a+b)(a^2ab+b^2)\] difference of cubes formula \[(ab)(a^2+ab+b^2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the answer is C.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not really sure on this one. I was thinking either A. or C.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4ah I see we have to go backwards on this.. so we need all perfect cubes. . no it can't be a.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because it can't be factored right? @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.48 27 64 \[2^3 = 8 , 3^3 =27, 4^3 = 64, 5^3 = 125\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4@200205650 true and also 14 is not a perfect cube

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4sum of cubes \[(a^3+b^3) = (a+b)(a^2ab+b^2) \] difference of cubes \[(a^3b^3) = (ab)(a^2+ab+b^2) \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4these choices are wonky though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't see how you can factor any of them in the answer choices...unless you use the quadratic formula.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4like .. let a = x let b = 4 \[(a^3b^3) = (ab)(a^2+ab+b^2) \] \[(x^34^3) = (x4)(x^2+4x+16) \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4I see perfect squares for the last 3 choices that's true.. for the ending part is b^2 . It's like this question isn't following the rules. That's what's getting me mad

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4let a = x and let b = 9 difference of cubes formula \[(a^3b^3) = (ab)(a^2+ab+b^2) \] \[(x^39^3) = (x9)(x^2+9x+81) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From your examples so far @UsukiDoll I would think C. is the right answer.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4that is messed up... really. 9 isn't a perfect cube to begin with. perfect square yes but not perfect cube

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya it isn't...but the way the equations are set up B. and D. don't follow that same pattern.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4only perfect cubes are allowed for these formulas...>:/ oh I'm missing something XD 9 x 9 x 9 = 729

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4729 is a perfect cube number. that guy is 9

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4ok I got it.. I missed a very important step let a = x and b = 9 \[(x^3729) = (x9)(x^2+9x+81)\] \[(x^39^3) = (x9)(x^2+9x+81)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is C.?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4that 729 is rewritten as 9^3 in the second line so a = x (the variable) and b = 9 (the number) a^2 = x^2 ab = 9x b^2 = (9)(9) = 81

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4yeah C matches ... on top of that .. the signs match for the difference of cubes. I wasn't working backwards until the very beginning.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great job explaining that!

lokiando
 one year ago
Best ResponseYou've already chosen the best response.0well thank you guys for your time and patients :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4XD patience XDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.4It's past my lunch time too... so that's why my thinking started to go off
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