Multiply out and simplify (x + 1)(x² − x + 1).

Step-by-step solution

1. 1

   Distribute the binomial across the trinomial. Split $(x+1)$ into its two terms and multiply each by the whole second factor:

   $(x+1)(x^2-x+1)=x(x^2-x+1)+1\cdot(x^2-x+1)$

2. 2

   Expand each piece.

   $x(x^2-x+1)=x^3-x^2+x$

   $1\cdot(x^2-x+1)=x^2-x+1$

3. 3

   Add and watch the middle terms annihilate.

   $x^3-x^2+x+x^2-x+1$

   The $-x^2$ and $+x^2$ cancel, and so do $+x$ and $-x$. Four of the six terms disappear:

   $x^3+1$

4. 4

   Recognise why the cancellation was guaranteed. This is the sum-of-cubes identity read backwards:

   $a^3+b^3=(a+b)\left(a^2-ab+b^2\right)$

   With $a=x$ and $b=1$ the second factor is $x^2-x\cdot 1+1^2=x^2-x+1$, exactly what was given. The alternating signs in $a^2-ab+b^2$ are precisely what makes the cross terms cancel — the factor is not $(x^2+x+1)$, which would instead pair with $(x-1)$ to give $x^3-1$.

5. 5

   Verify with a value. At $x=2$: the original is $(3)(4-2+1)=3\cdot 3=9$, and $x^3+1=8+1=9$. ✓ At $x=-1$ both sides give $0$, consistent with $x=-1$ being the only real root of $x^3+1$.