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 answers are correct is shown below.


\[\begin{array}{c} \eqalign{ & {\left( {{x_1}} \right)^3} = {\left( {2 + 0\,i} \right)^3} \cr & {\left( {{x_1}} \right)^3} = \left( 2 \right)\left( 2 \right)\left( 2 \right) \cr & {\left( {{x_1}} \right)^3} = 8 \cr} \end{array}\]
\[\begin{array}{c} \eqalign{ & {\left( {{x_2}} \right)^3} = {\left( { – 1 + \sqrt 3 \,i} \right)^3} \cr & {\left( {{x_2}} \right)^3} = \left( { – 1 + \sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( {\left( { – 1} \right)\left( { – 1} \right) + \left( { – 1} \right)\left( {\sqrt 3 \,i} \right) + \left( {\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( {\sqrt 3 \,i} \right)\left( {\sqrt 3 \,i} \right)} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( {1 – \sqrt 3 \,i – \sqrt 3 \,i – 3} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( { – 2 – 2\sqrt 3 \,i} \right)\left( { – 1 + \sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = \left( { – 2} \right)\left( { – 1} \right) + \left( { – 2} \right)\left( {\sqrt 3 \,i} \right) + \left( { – 2\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( { – 2\sqrt 3 \,i} \right)\left( {\sqrt 3 \,i} \right) \cr & {\left( {{x_2}} \right)^3} = 2 – 2\sqrt 3 \,i + 2\sqrt 3 \,i + 6\, \cr & {\left( {{x_2}} \right)^3} = 2 + 6 \cr & {\left( {{x_2}} \right)^3} = 8 \cr} \end{array}\]
\[\begin{array}{c} \eqalign{ & {\left( {{x_3}} \right)^3} = {\left( { – 1 – \sqrt 3 \,i} \right)^3} \cr & {\left( {{x_3}} \right)^3} = \left( { – 1 – \sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( {\left( { – 1} \right)\left( { – 1} \right) + \left( { – 1} \right)\left( { – \sqrt 3 \,i} \right) + \left( { – \sqrt 3 \,i} \right)\left( { – 1} \right) + \left( { – \sqrt 3 \,i} \right)\left( { – \sqrt 3 \,i} \right)} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( {1 + \sqrt 3 \,i + \sqrt 3 \,i – 3} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( { – 2 + 2\sqrt 3 \,i} \right)\left( { – 1 – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = \left( { – 2} \right)\left( { – 1} \right) + \left( { – 2} \right)\left( { – \sqrt 3 \,i} \right) + \left( {2\sqrt 3 \,i} \right)\left( { – 1} \right) + \left( {2\sqrt 3 \,i} \right)\left( { – \sqrt 3 \,i} \right) \cr & {\left( {{x_3}} \right)^3} = 2 + 2\sqrt 3 \,i – 2\sqrt 3 \,i + 6\, \cr & {\left( {{x_3}} \right)^3} = 2 + 6 \cr & {\left( {{x_3}} \right)^3} = 8 \cr} \end{array}\]