# Solution to Practice Problem #2

There is always one oneth root, two square roots, three cube roots, etc.

So, there are three cube roots of 8, and they are in the complex plane.

$\begin{array}{c} \root 3 \of 8 \end{array}$

\begin{array}{c} \eqalign{ & {x_1} = \,\,\,\,2 \,+ \,\,\,0\,i \cr & {x_2} = \,\, – 1 \,+ \sqrt 3 \,i \cr & {x_3} = \,\, – 1 \,\,\, – \, \sqrt 3 \,i \cr} \end{array}

This is an advanced topic for High School Precalculus and the SAT Math Level-2 Subject Test, and lessons are necessary to properly teach how to arrive at the answers. Verifying that these three roots are correct is shown below.

Checking the first of the three roots:

\begin{array}{c} \eqalign{ & {\left( {{x_1}} \right)^3} = {\left( {2 + 0\,i} \right)^3} \cr & {\left( {{x_1}} \right)^3} = \left( 2 \right)\left( 2 \right)\left( 2 \right) \cr & {\left( {{x_1}} \right)^3} = 8 \cr} \end{array}

Checking the second of the three roots:

\begin{array}{c} \eqalign{ & {\left( {{x_2}} \right)^3} = {\left( { – 1 + \sqrt 3 \,i} \right)^3} \cr & {\left( {{x_2}} \right)^3} = \left( { – 1 + \sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( {\left( { – 1} \right)\left( { – 1} \right) + \left( { – 1} \right)\left( {\sqrt 3 \,i} \right) + \left( {\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( {\sqrt 3 \,i} \right)\left( {\sqrt 3 \,i} \right)} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( {1 – \sqrt 3 \,i – \sqrt 3 \,i – 3} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( { – 2 – 2\sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( { – 2} \right)\left( { – 1} \right) + \left( { – 2} \right)\left( {\sqrt 3 \,i} \right) + \left( { – 2\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( { – 2\sqrt 3 \,i} \right)\left( {\sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = 2 – 2\sqrt 3 \,i + 2\sqrt 3 \,i + 6\, \cr & {\left( {{x_2}} \right)^3} = 2 + 6 \cr & {\left( {{x_2}} \right)^3} = 8 \cr} \end{array}

Checking the third of the three roots:

\begin{array}{c} \eqalign{ & {\left( {{x_3}} \right)^3} = {\left( { – 1 – \sqrt 3 \,i} \right)^3} \cr & {\left( {{x_3}} \right)^3} = \left( { – 1 – \sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( {\left( { – 1} \right)\left( { – 1} \right) + \left( { – 1} \right)\left( { – \sqrt 3 \,i} \right) + \left( { – \sqrt 3 \,i} \right)\left( { – 1} \right) + \left( { – \sqrt 3 \,i} \right)\left( { – \sqrt 3 \,i} \right)} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( {1 + \sqrt 3 \,i + \sqrt 3 \,i – 3} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( { – 2 + 2\sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( { – 2} \right)\left( { – 1} \right) + \left( { – 2} \right)\left( { – \sqrt 3 \,i} \right) + \left( {2\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( {2\sqrt 3 \,i} \right)\left( { – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = 2 + 2\sqrt 3 \,i – 2\sqrt 3 \,i + 6\, \cr & {\left( {{x_3}} \right)^3} = 2 + 6 \cr & {\left( {{x_3}} \right)^3} = 8 \cr} \end{array}