Solve each equation.
47. [tex]$x^6+7 x^3-8=0$[/tex]

Substitute y = x 3 to transform the equation into a quadratic equation: y 2 + 7 y − 8 = 0 .
Factor the quadratic equation: ( y + 8 ) ( y − 1 ) = 0 , which gives y = − 8 or y = 1 .
Solve x 3 = − 8 and x 3 = 1 to find the six roots of the original equation.
The solutions are x = 1 , − 2 , − 2 1 ​ + i 2 3 ​ ​ , − 2 1 ​ − i 2 3 ​ ​ , 1 + i 3 ​ , 1 − i 3 ​ .

Explanation

Problem Analysis We are asked to solve the equation x 6 + 7 x 3 − 8 = 0 . This is a sixth-degree polynomial equation. However, we can simplify it by recognizing that it is a quadratic equation in disguise.

Substitution Let y = x 3 . Then the equation becomes y 2 + 7 y − 8 = 0 . This is a quadratic equation in terms of y .

Factoring We can solve this quadratic equation by factoring. We look for two numbers that multiply to -8 and add to 7. These numbers are 8 and -1. Thus, we can factor the quadratic equation as ( y + 8 ) ( y − 1 ) = 0 .

Solving for y The solutions for y are y = − 8 and y = 1 .

Substituting Back Since y = x 3 , we have x 3 = − 8 and x 3 = 1 . Now we need to solve these two cubic equations for x .

Solving x^3 = -8 First, let's solve x 3 = − 8 . We can rewrite this as x 3 + 8 = 0 . We know that x = − 2 is one real solution since ( − 2 ) 3 = − 8 . The other two solutions are complex. We can express -8 as 8 e iπ . Then the cube roots of -8 are 2 e i ( π /3 + 2 πk /3 ) for k = 0 , 1 , 2 . These are:


For k = 0 : 2 e iπ /3 = 2 ( cos ( π /3 ) + i sin ( π /3 )) = 2 ( 2 1 ​ + i 2 3 ​ ​ ) = 1 + i 3 ​ .
For k = 1 : 2 e iπ = 2 ( cos ( π ) + i sin ( π )) = 2 ( − 1 + 0 i ) = − 2 .
For k = 2 : 2 e i 5 π /3 = 2 ( cos ( 5 π /3 ) + i sin ( 5 π /3 )) = 2 ( 2 1 ​ − i 2 3 ​ ​ ) = 1 − i 3 ​ .
So the solutions to x 3 = − 8 are x = − 2 , x = 1 + i 3 ​ , and x = 1 − i 3 ​ .

Solving x^3 = 1 Next, let's solve x 3 = 1 . We can rewrite this as x 3 − 1 = 0 . We know that x = 1 is one real solution since 1 3 = 1 . The other two solutions are complex. We can express 1 as 1 e i 0 . Then the cube roots of 1 are 1 e i ( 0 + 2 πk ) /3 for k = 0 , 1 , 2 . These are:

For k = 0 : e i 0 = cos ( 0 ) + i sin ( 0 ) = 1 + 0 i = 1 .
For k = 1 : e i 2 π /3 = cos ( 2 π /3 ) + i sin ( 2 π /3 ) = − 2 1 ​ + i 2 3 ​ ​ .
For k = 2 : e i 4 π /3 = cos ( 4 π /3 ) + i sin ( 4 π /3 ) = − 2 1 ​ − i 2 3 ​ ​ .
So the solutions to x 3 = 1 are x = 1 , x = − 2 1 ​ + i 2 3 ​ ​ , and x = − 2 1 ​ − i 2 3 ​ ​ .

Listing All Solutions Therefore, the six solutions for x are x = 1 , − 2 , − 2 1 ​ + i 2 3 ​ ​ , − 2 1 ​ − i 2 3 ​ ​ , 1 + i 3 ​ , 1 − i 3 ​ .

Final Answer The solutions to the equation x 6 + 7 x 3 − 8 = 0 are:


x = 1 x = − 2 x = − 2 1 ​ + i 2 3 ​ ​ x = − 2 1 ​ − i 2 3 ​ ​ x = 1 + i 3 ​ x = 1 − i 3 ​
Examples
Understanding polynomial equations and their solutions is crucial in many fields, including engineering, physics, and computer science. For instance, when designing a bridge, engineers use polynomial equations to model the forces and stresses acting on the structure. Solving these equations helps them determine the bridge's stability and safety. Similarly, in computer graphics, polynomial equations are used to create realistic images and animations. By finding the roots of these equations, developers can accurately render shapes, shadows, and textures, making virtual environments more immersive and visually appealing.

Answered by GinnyAnswer | 2025-07-03

We solved the equation x 6 + 7 x 3 − 8 = 0 by substituting y = x 3 , turning it into a quadratic equation. The factorized form led us to the solutions for y , which we back-substituted to find the six solutions for x . The final solutions are x = 1 , − 2 , − 2 1 ​ + i 2 3 ​ ​ , − 2 1 ​ − i 2 3 ​ ​ , 1 + i 3 ​ , 1 − i 3 ​ .
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Answered by Anonymous | 2025-07-04