At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
sum of cubes formula \[(a+b)(a^2-ab+b^2)\] difference of cubes formula \[(a-b)(a^2+ab+b^2)\]
will give medal
I think the answer is C.
you think? mmmmm
Not really sure on this one. I was thinking either A. or C.
ah I see we have to go backwards on this.. so we need all perfect cubes. . no it can't be a.
Because it can't be factored right? @UsukiDoll
8 27 64 \[2^3 = 8 , 3^3 =27, 4^3 = 64, 5^3 = 125\]
@200205650 true and also 14 is not a perfect cube
so its c?
sum of cubes \[(a^3+b^3) = (a+b)(a^2-ab+b^2) \] difference of cubes \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \]
these choices are wonky though
I don't see how you can factor any of them in the answer choices...unless you use the quadratic formula.
like .. let a = x let b = 4 \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-4^3) = (x-4)(x^2+4x+16) \]
I 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
let a = x and let b = 9 difference of cubes formula \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-9^3) = (x-9)(x^2+9x+81) \]
From your examples so far @UsukiDoll I would think C. is the right answer.
that is messed up... really. 9 isn't a perfect cube to begin with. perfect square yes but not perfect cube
Ya it isn't...but the way the equations are set up B. and D. don't follow that same pattern.
only perfect cubes are allowed for these formulas...>:/ oh I'm missing something XD 9 x 9 x 9 = 729
729 is a perfect cube number. that guy is 9
ok I got it.. I missed a very important step let a = x and b = 9 \[(x^3-729) = (x-9)(x^2+9x+81)\] \[(x^3-9^3) = (x-9)(x^2+9x+81)\]
So the answer is C.?
that 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
yeah C matches ... on top of that .. the signs match for the difference of cubes. I wasn't working backwards until the very beginning.
Great job explaining that!
well thank you guys for your time and patients :)
XD patience XDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
It's past my lunch time too... so that's why my thinking started to go off