Help solving college algebra problem?

Question

Im having issues solving for x with this problem. I cant seem to find a way to get an answer

(x+3)^3=(-48x^2)/6x^5

I have been trying multiple ways of solving it and have yet to get a correct solution.

Answer 1

The equation can be simplified as follows:

(x + 3)^3 = (-48x^2) / 6x^5.
(x + 3)^3 = -8 / x^3. (-48 / 6 = -8, x^2 / x^5 = 1 / x^3)
x + 3 = -2 / x. (Take cube root of each side)
x^2 + 3x = -2. (Multiply each side by x)
x^2 + 3x + 2 = 0. (Add 2 to each side)
(x + 2) (x + 1) = 0. (Factor remaining equation)
x = -1 or -2.

Answer 2

The right hand side simplifies to -8/x^3

So we have (x+3)^3 = -8/x^3

Hey, let's take the cube root of both sides!

(x+3) = -2/x

Let's multiply by x

x^2 + 3x = -2

x^2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x is -2 or x is -1

Answer 3

at the start take LCM of the given question. u might get denominator as (x-14)(x+14) (3(x+14)+3(x-14))/((x-14)(x+14)) = 10/(x^2 - 196) right here u might might desire to prepare the formulation of (a+b)(a-b) = a^-b^2 (3(x+14)+3(x-14))/(x^2-196)=10/(x^2-19... now right here do go multiplication to locate the respond (3x+40 two+3x-40 two)(x^2-196) = 10(x^2-196) 6x = 10 sine x^2-196 may be shrink u might get x=10/6 x=5/3 hence ur answer is 5/3 desire u have been given the answer if any issue u can tell me

Answer 4

(x+3)^3=(-48x^2)/6x^5
(x+3)^3=-8x^(-3)
(x+3)^3=1/-24x

sorry that's as far as i could i get, i'm only in yr 9...lol...but i hope it helps anyway...