$\text{If}x+y+z=1,xy+yz+zx=-1\text{and}xyz=-1,\text{then}(x^3+y^3+z^3)\text{is equal to}\text{\_\_\_\_\_}.$

The correct answer is 1.

To find the value of $x^3+y^3+z^3$, we use the standard algebraic identity relating the sum of cubes to the sum of variables, the sum of pairwise products, and their product:

$x^3+y^3+z^3-3xyz=(x+y+z)\left[\right(x^2+y^2+z^2)-(xy+yz+zx\left)\right]$

Step 1: Calculate $x^2+y^2+z^2$

We start with the expansion formula for the square of a trinomial:

$(x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$

Rearranging the terms to solve for $x^2+y^2+z^2$:

$x^2+y^2+z^2=(x+y+z{)}^2-2(xy+yz+zx)$

Substitute the given values $x+y+z=1$ and $xy+yz+zx=-1$ into the equation:

$x^2+y^2+z^2=(1{)}^2-2(-1)$

$x^2+y^2+z^2=1+2=3$

Step 2: Evaluate $x^3+y^3+z^3$

Now substitute $x+y+z=1$, $xy+yz+zx=-1$, $xyz=-1$, and $x^2+y^2+z^2=3$ into the main algebraic identity:

$x^3+y^3+z^3-3(-1)=\left(1\right)[3-(-1\left)\right]$

$x^3+y^3+z^3+3=1\times (3+1)$

$x^3+y^3+z^3+3=4$

$x^3+y^3+z^3=4-3$

$x^3+y^3+z^3=1$

Thus, $x^3+y^3+z^3$ is equal to 1.