r^3+216/r+6=
r ^ 3 + 216 / r + 6 = r ^ 3 * r / r + 216 / r + 6 r / r =
( r ^ 4 + 216 + 6 r )/ r=
( r ^ 4 + 6 r + 216 ) / r
I am very sure you meant:
(r^3+216)/(r+6)
= (r+6)(r^2 - 6r + 36)/(r+6)
= r^2 - 6r + 36
To solve the equation r^3 + 216/r + 6 = 0, we can use a common technique called factoring.
Step 1: Bring all terms to one side of the equation to obtain: r^3 + 6r + 216 = 0.
Step 2: Notice that the equation can be rewritten as a sum of cubes. The equation r^3 + 6r + 216 = 0 can be rewritten as (r^3 + 6r + 9^3) = 0.
Step 3: Now we have a sum of cubes, which can be factored using the following formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2). In this case, a = r and b = 3.
So, applying the formula, we get (r + 3)((r)^2 - (r)(3) + (3)^2) = 0.
Simplifying this further, (r + 3)(r^2 -3r + 9) = 0.
We now have two factors that can individually equal zero to solve for r:
Factor 1: r + 3 = 0 --> r = -3
Factor 2: r^2 -3r + 9 = 0
To solve the quadratic equation, we can employ either factoring, completing the square, or the quadratic formula. However, in this case, the quadratic equation does not factor nicely, and completing the square can be a bit tedious. Therefore, we will use the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / 2a.
For the quadratic r^2 -3r + 9 = 0, we have a = 1, b = -3, and c = 9. Substituting these values into the quadratic formula:
r = (-(-3) ± √((-3)^2 - 4(1)(9))) / (2(1))
= (3 ± √(9 - 36)) / 2
= (3 ± √(-27)) / 2.
Since we cannot take the square root of a negative number in the real number system, this equation has no real solutions for r.
Therefore, the solutions to the equation r^3 + 216/r + 6 = 0 are r = -3 and there are no real solutions for r when considering the quadratic factor.