what is the factored form of $x^{3}-1$?\n$(x^{3}-1)(x^{2}+x + 1)$\n$(x - 1)(x^{2}-x + 1)$\n$(x - 1)(x^{2}+x…

what is the factored form of $x^{3}-1$?\n$(x^{3}-1)(x^{2}+x + 1)$\n$(x - 1)(x^{2}-x + 1)$\n$(x - 1)(x^{2}+x + 1)$\n$(x^{3}-1)(x^{2}+2x + 1)$

what is the factored form of $x^{3}-1$?\n$(x^{3}-1)(x^{2}+x + 1)$\n$(x - 1)(x^{2}-x + 1)$\n$(x - 1)(x^{2}+x + 1)$\n$(x^{3}-1)(x^{2}+2x + 1)$

Answer

Explanation:

Step1: Recall the difference - of - cubes formula

The difference - of - cubes formula is $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$. In the expression $x^{3}-1$, we have $a=x$ and $b = 1$.

Step2: Substitute values into the formula

Substituting $a=x$ and $b = 1$ into the formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$, we get $x^{3}-1=(x - 1)(x^{2}+x\times1+1^{2})=(x - 1)(x^{2}+x + 1)$.

Answer:

C. $(x - 1)(x^{2}+x + 1)$