Sum/Difference of Cubes
When solving a sum of cubes or a difference of cubes you will always get one real soloution and 2 complex conjugate solutions.
For the two complex solutions, you need to solve the trinomial using the Quadratic Formula.
Prove that when using the quadratic formula for the two complex solutions, the radicand in the final answer is always 3 (sqrt3). In addition, prove that the value of the '-b' and the coefficient of the square root is always the same number.
HINT: This is an algebraic proof. Use a general example that will apply to all cases. You only need to show it for the sum of cubes or difference of cubes, but not both.
(a^3 x^3 - b^3) = (ax-b)(a^2 x^2 + abx + b^2)
(a^3 x^3 + b^3) = (ax+b)(a^2 x^2 - abx + b^2)
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You need to solve for x the equation`a^3x^3 + b^3 = 0` , hence, you need to convert the sum of cubes into a product such that:
`a^3x^3 + b^3 = (ax + b)(a^2x^2- abx + b^2)`
`a^3x^3 + b^3 = 0 => (ax + b)(a^2x^2 - abx + b^2) = 0`
You need to set the factors of product equal to zero such that:
`ax + b = 0 => ax = -b => x = -b/a`
`a^2x^2 - abx + b^2 = 0`
You need to use quadratic formula such that:
`x_(1,2) = (ab+-sqrt(a^2b^2 - 4a^2b^2))/(2a^2)`
`x_(1,2) = (ab+-sqrt(-3a^2b^2))/(2a^2)`
Since `sqrt -1 = i ` (complex number theory), yields:
`x_(1,2) = (ab+-ab*sqrt3*i)/(2a^2)`
Factoring out a yields:
`x_(1,2) = (a(b+-b*sqrt3*i))/(2a^2) => x_(1,2) = (b+-b*sqrt3*i)/(2a) `
Hence, evaluating the roots of the equation `a^3x^3 + b^3 = 0` yields one real root `x = -b/a` and two complex conjugate roots `x_(1,2) = (b+-b*sqrt3*i)/(2a).`
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