Resolve $a^{3} – b^{3} + 1 +3ab$ into factors.

Given: $a^{3} – b^{3} + 1 +3ab$.

To do: To factorize: $a^{3} – b^{3} + 1 +3ab$.

Solution:

$a^{3} – b^{3} + 1 +3ab$

$=(a)^3+(-b)^3+1+3\times a\times b\times 1$

On applying [$x^3+y^3+z^3=x^2+y^2+z^2-xy-yz-zx$]

$=( a+( -b)+1)( a^2+(-b)^2+1^2-a( -b)-( -b)(1)-1(a))$

$=( a-b+1)( a^2+b^2+1+ab+b-a)$

 

Thus, $a^{3} – b^{3} + 1 +3ab=( a-b+1)( a^2+b^2+1+ab+b-a)$.

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Updated on: 10-Oct-2022

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