## Precalculus (6th Edition) Blitzer

The result in the standard form is $8$.
Perform the standard operation and expand the given expression as below: $\left( -1+i\sqrt{3} \right)\left( -1+i\sqrt{3} \right)\left( -1+i\sqrt{3} \right)$. \begin{align} & =\,\left( -1+i\sqrt{3} \right)\left( 1-i\sqrt{3}-i\sqrt{3}+3{{i}^{2}} \right) \\ & =\left( -1+i\sqrt{3} \right)\left( 1-2i\sqrt{3}+3{{i}^{2}} \right) \\ & =\left( -1+i\sqrt{3} \right)\left( 1-2i\sqrt{3}-3 \right) \\ & =\left( -1+i\sqrt{3} \right)\left( -2-2i\sqrt{3} \right) \end{align} Now, perform the multiplication operation again on the above reduced equation and simplify it as below: \begin{align} & =\left( -1+i\sqrt{3} \right)\left( -2-2i\sqrt{3} \right) \\ & =\left( 2+2i\sqrt{3}-2i\sqrt{3}-6{{i}^{2}} \right) \\ & =2+6 \\ & =8 \\ \end{align} Therefore, the result in the standard form is $8$.