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Rebbbtx

  • 3 years ago

Can someone please show me step by step, how to factor this equation? a^3x-b^3x-a^3y+b^3y the answer = (a-b)(a^2+ab+b^2)(x-y)

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  1. sauravshakya
    • 3 years ago
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    a^3x-b^3x-a^3y+b^3y (a^3-b^3)x - (a^3 -b^3)y

  2. sauravshakya
    • 3 years ago
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    Did u get that step?

  3. Rebbbtx
    • 3 years ago
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    okay, yes. First you take the x and y out

  4. Rebbbtx
    • 3 years ago
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    what next?

  5. Hezekieli
    • 3 years ago
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    One could also start by taking a and b out:\[a^{3}(x-y)+b^{3}(y-x)\] But I don't really know how to get to that answer or why to aim at that? :o

  6. andriod09
    • 3 years ago
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    The best way to do it: FIND THE COMMON NUMBER BETWEEN EVERYTHING.

  7. abdul_shabeer
    • 3 years ago
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    \[a^3-b^3=(a-b)(a^2+ab+b^2)\]

  8. abdul_shabeer
    • 3 years ago
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    Got the answer?

  9. Hezekieli
    • 3 years ago
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    You can check that what abdul said by going backwards: \[(a - b)(a ^{2} + ab + b ^{2})\] \[= (a^{3}+a^{2}b + ab^{2}) + (-a^{2}b - ab^{2} - b^{3})\] \[= a^{3}+a^{2}b - a^{2}b + ab^{2} - ab^{2} - b^{3}\] \[= a^{3} - b^{3}\] Not sure yet how to get that (x - y) out o.O

  10. abdul_shabeer
    • 3 years ago
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    \[a^3x-b^3x-a^3y+b^3y\] \[x(a^3-b^3)-y(a^3-b^3)\] \[(a^3-b^3)(x-y)\] \[(a-b)(a^2+ab+b^2)(x-y)\]

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