Question:
Find the value
$a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$
Solution:
$=(a-b)^{3}+23$
$\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$
$=(a-b+2)\left((a-b)^{2}-(a-b)^{2}+2^{2}\right)$
$\therefore\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2(a-b)+4\right)$
$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$
$\therefore a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$
$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$
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