R3 – Cube Roots

Similar to R3 Division, cube roots in R3 are solved using simultaneous equations.

Simultaneous Equations Needed

The following is a method for finding R3 cube roots using simultaneous equations. For example, take the cubic equation:

• (3^v + 4i^{^} + 5j^v )³ = (3^v + 4i^{^} + 5j^v )*(3^v + 4i^{^} + 5j^v )²

As described in an earlier post, the square of the expression (a^? + bi^? + cj^?)²  is (a^2? +  b²i² + c²j²). This allows the calculation of (3^v + 4i^{^} + 5j^v )³ as follows:

• (3^v + 4i^{^} + 5j^v )³ = (3^v + 4i^{^} + 5j^v )*(3^v + 4i^{^} + 5j^v )²
  • = (3^v + 4i^{^} + 5j^v )*(9^v + 16i^{^} + 25j^v )
  • = (47^v + 34i^{^} + 137j^v )
  • => (3^v + 4i^{^} + 5j^v )³ = (47^v + 34i^{^} + 137j^v )
  • => ³√(47^v + 34i^{^} + 137j^v ) = (3^v + 4i^{^} + 5j^v )

Next, to find the cube root of (47^v + 34i^{^} + 137j^v ) solve the following equation:

• (a^? + bi^? + cj^?)*(a^? + bi^? + cj^?)*(a^? + bi^? + cj^?) = (47^v + 34i^{^} + 137j^v )

The method that follows uses logic and simultaneous equations to work out this cube root.

Method for Finding R3 Cube Roots

Generating the Simultaneous Equations

Multiplying out (a^? + bi^? + cj^?)³ gives the following result.

• (a^? + bi^? + cj^?)³ = (a^3? + bi^?*c²j^? + cj^?*b²i^?) + (b³i^?+ a^?*c²j^?  + cj^?*a^2?) + (c³j^? +a^?*b²i^? + bi^?*a^2?)

The multiplication table allows the gathering of the terms for each Opposite Type within brackets. Consequently, the generation of 3 equations for the example ³√(47^v + 34i^{^} + 137j^v ) is possible:

• A: (a^3? + bi^?*c²j^? + cj^?*b²i^?) = 47^v
• B: (b³i^?+ a^?*cj^2?  + cj^?*a^2?) = 34i^{^}
• C: (c³j^? +a^?*bi^2? + bi^?*a ^2?) = 137j^v

Simultaneous equations work purely with Counters as the Opposite Signs are unknown. The multiplication table shows that the cube of an Opposite Value such as a^3? leaves the Opposite Sign of the result the same, so treat a^3? as a³ for the simultaneous equation.

The Opposite Sign of the products of a bi^?*c²j^? and the product of cj^?*b²i^? are ^{^} or ^v. As a result of these 2 possibilities, several different simultaneous equations require testing for an answer as follows:

• a³ + b*c² + c*b² + 47 = 0
• a³ + b*c² + ^–c*b² + 47 = 0
• a³ + ^–b*c² + c*b² + 47 = 0
• a³ +^–b*c² + ^–c*b² + 47 = 0

Similarly, Equation B generates the following possible simultaneous equations:

• b³ + a*c²  + c*a² + ^–34 = 0
• b³ + a*c²  + ^–c*a² + ^–34 = 0
• b³ + ^–a*c²  + c*a² + ^–34 = 0
• b³ + ^–a*c²  + ^–c*a² + ^–34 = 0

Equation C generates the following possible simultaneous equations:

• c³ + a*b² + b*a² + 137 = 0
• c³ + a*b² + ^–b*a² + 137 = 0
• c³ + ^–a*b² + b*a² + 137 = 0
• c³ + ^–a*b² + ^–b*a² + 137 = 0

Solving the simultaneous equations

Attempts are made to solve each combination of the Equation A, B and C equations, but solutions for R3 cube roots are not always found. However, any solution found is tested to see if it is a valid cubic root. This is done by cubing the solution.

The equations that show the correct R3 cube root solution of (3^v + 4i^{^} + 5j^v ) for ³√(47^v + 34i^{^} + 137j^v ) follow. As the simultaneous equations are dealing with Counters, this is reflected by a flip sign for ^v signed values such as 3^v and 5j^v:

• a³ + ^–b*c² + ^–c*b² + 47 = 0
  • => ^–3³ + ^–4*^–5² + ^{– –}5*4² + 47 = 0
  • => ^–27 + ^–100 + 80 + 47 = 0
  • => 0 = 0

• b³ + a*cj²  + ^–c*a² + ^–34 = 0
  • => 4³ + ^–3*^–5² + ^{– –}5*^–3² + ^–34 = 0
  • => 64 + ^–75 + 45 + ^–34 = 0
  • => 0 = 0

• c³ + a*b² + b*a² + 137^{^} = 0
  • => ^–5³ + ^–3*4² + 4*^–3² + 137 = 0
  • => ^–125 + ^–48 + 36 + 137 = 0
  • => 0 = 0

Next: Exponentiation

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