Question
Factorise:
$x^3-3 x^2+3 x+7$
$x^3-3 x^2+3 x+7$
✓
Answer
$x^3-3 x^2+3 x+7$
$=x^3-3 x^2+3 x-1+8$
$=\left(x^3-3 x^2+3 x-1\right)+8$
$=(x-1)^3+2^3$
$=(x-1+2)\left[(x-1)^2-(x-1)(2)+2^2\right]$
$=(x+1)\left(x^2-2 x+1-2 x+2+4\right)$
$=(x+1)\left(x^2-4 x+7\right)$
$=x^3-3 x^2+3 x-1+8$
$=\left(x^3-3 x^2+3 x-1\right)+8$
$=(x-1)^3+2^3$
$=(x-1+2)\left[(x-1)^2-(x-1)(2)+2^2\right]$
$=(x+1)\left(x^2-2 x+1-2 x+2+4\right)$
$=(x+1)\left(x^2-4 x+7\right)$
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