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24 December, 00:45

Prove that

a³+b³ = (a+b) ³-3ab (a+b)

a³-b³ = (a-b) ³+3ab (a-b)

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  1. 24 December, 02:44
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    (For the first proof)

    Start with (a+b) ^3

    (a+b) ^3 = ((a+b) ^2) * (a+b)

    or, (a+b) ^3 = (a^2+b^2+2ab) * (a+b)

    or, (a+b) ^3=a^3+b*a^2+b^2*a+b^3+2a^2*b+2ab^2

    or, (a+b) ^3=a^3+b^3+3a^2*b+3a*b^2

    or, (a+b) ^3=a^3+b^3+3ab (a+b)

    or, a^3+b^3 = (a+b) ^3-3ab (a+b)

    (For the second proof)

    Start with (a-b) ^3

    (a-b) ^3 = ((a-b) ^2) * (a-b)

    or, (a-b) ^3 = (a^2+b^2-2ab) * (a-b)

    or, (a-b) ^3

    =a^3-b*a^2+b^2*a-b^3-2a^2*b+2ab^2

    or, (a-b) ^3=a^3-b^3-3a^2*b+3a*b^2

    or, (a-b) ^3=a^3-b^3-3ab (a-b)

    or, a^3-b^3 = (a-b) ^3+3ab (a-b)
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