Question 166923
{{{(s-6)^3=27(s-8)}}} Start with the given equation



{{{s^3-18s^2+108s-216 =27(s-8)}}} Expand {{{(s-6)^3}}} to get {{{s^3-18s^2+108s-216}}}. Hint: Use Pascal's Triangle to expand



{{{s^3-18s^2+108s-216 = 27s - 216}}} Distribute



{{{s^3-18s^2+108s = 27s}}} Add 216 to both sides. Notice that they cancel out and go away.



{{{s^3-18s^2+108s - 27s = 0}}} Subtract 27s from both sides.



{{{s^3-18s^2+ 81s = 0}}} Combine like terms.



{{{s(s^2-18s+81)=0}}} Factor out the GCF "s" from the left side



{{{s(s-9)(s-9)=0}}} Factor {{{s^2-18s+81}}} to get {{{(s-9)(s-9)}}}



Now set each factor equal to zero:



{{{s=0}}}, {{{s-9=0}}} or {{{s-9=0}}}



{{{s=0}}}, {{{s=9}}} or {{{s=9}}}   Now solve for s in each case



Since we have a repeating answer, our answers are {{{s=0}}} or  {{{s=9}}} 



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Answer:


So the solutions are {{{s=0}}} or {{{s=9}}} where {{{s=9}}} is a double root (ie has a multiplicity of 2)