Find the real solution(s) of the equation 1/6x^3 = -36

x3 - 36x = 16(x-6) => x(x2 - 36) = 16(x-6) [notice on the left we now have the difference of two squares]

                            => x(x - 6)(x+6) = 16(x-6)

                            =>x(x - 6)(x+6) - 16(x - 6) = 0 [now we can factor x-6 from both terms]

                            => (x - 6)(x(x + 6) -16) = 0

                            => (x- 6)(x2 + 6x - 16) = 0

                            => (x - 6)(x + 8)(x - 2) = 0

So now that we have our factorization, our solutions are the values of x which make this equation true. 

That is x = 6, x = -8, and x = 2

Below.

Step-by-step explanation:

1/6x^3 = -36

Multiply both sides by 6:

x^3 = -216

x = ∛(-216)

= -6.

after going and makeing graphs i would go with a,

there is nothing that is matches the graph. if there was i would show you. but therefore my answer is a.

hope i

-lavira: )

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