Resolve a³-b³+1+3ab into factors
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(a+b)³ is your question answer
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a^3-b^3+1+3ab
=a^b+(-b)^3+1^3-3a(-b)c
Using the formula,
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2^c^2-ab-bc-ca)
U can find your factors which will be
(a-b+1)(a^2+(-b)^2+1+ab+bc-a)
=a^b+(-b)^3+1^3-3a(-b)c
Using the formula,
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2^c^2-ab-bc-ca)
U can find your factors which will be
(a-b+1)(a^2+(-b)^2+1+ab+bc-a)
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