Question 80737
{{{8(c^3-c) = c (6c+1)}}}


{{{8c^3-8c = 6c^2+c}}} Distribute


{{{8c^3-6c^2-8c-c = 0}}} Get all terms to one side


{{{8c^3-6c^2-9c = 0}}} Combine like terms


{{{c(8c^2-6c-9) = 0}}} Factor out a c


Now factor {{{8c^2-6c-9}}}

note: this solver uses x instead of c
*[invoke quadratic_factoring 8, -6, -9]


So we have


{{{c(4c+3)(2c-3) = 0}}}

which gives us


{{{c=0}}} or {{{4c+3=0}}} or {{{2c-3=0}}} Set each individual factor equal to zero


{{{c=0}}} or {{{4c=-3}}} or {{{2c=3}}} Now solve for c in each case


So our answer is


{{{c=0}}} or {{{c=-3/4}}} or {{{c=3/2}}}