find the factors of a(power)3 - b(power)3 + 1 + 3ab into factors.
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a^3 -b^3 +(1)^3 + 3ab
Could be writen as
a^3 +(-b)^3 +(1)^3 - (3)(a)(-b)(1)
Using the identity:
a^3 + b^3 +c^3 -3abc which equals (a+b+c) (a^2+b^2 +c^2 -ab-bc-ca)
a^3 +(-b)^3 +(1)^3 - (3)(a)(-b)(1) would be factorised as
{a+(-b) +1} {a^2 +(-b)^2 +1^2 - (a)(-b)- (-b )(1) - a}
{a-b +1} {a^2 +b^2 +1 +ab+b -a}
Could be writen as
a^3 +(-b)^3 +(1)^3 - (3)(a)(-b)(1)
Using the identity:
a^3 + b^3 +c^3 -3abc which equals (a+b+c) (a^2+b^2 +c^2 -ab-bc-ca)
a^3 +(-b)^3 +(1)^3 - (3)(a)(-b)(1) would be factorised as
{a+(-b) +1} {a^2 +(-b)^2 +1^2 - (a)(-b)- (-b )(1) - a}
{a-b +1} {a^2 +b^2 +1 +ab+b -a}
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