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x3 divided by (x+1)
Consider x3 as x3+0x2+0x+0
and either do it by long division:
x²- x+1-
x+1)x³+0x²+0x+0
x³+ x²
-x²+0x
-x²- x
x+0
x+1
-1
Or by synthetic division:
-1|1 0 0 0
| -1 1 -1
1 -1 1 -1
also yielding: x²-x+1-
Edwin