Question 599123


{{{s^6-729}}} Start with the given expression.



{{{(s^2)^3-(9)^3}}} Rewrite {{{s^6}}} as {{{(s^2)^3}}}. Rewrite {{{729}}} as {{{(9)^3}}}.



{{{(s^2-9)((s^2)^2+(s^2)(9)+(9)^2)}}} Now factor by using the difference of cubes formula. Remember the <a href="http://www.purplemath.com/modules/specfact2.htm">difference of cubes formula</a> is {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}



{{{(s^2-9)(s^4+9s^2+81)}}} Multiply



{{{(s-3)(s+3)(s^4+9s^2+81)}}} Now factor {{{s^2-9}}} to get {{{(s-3)(s+3)}}} (using the difference of squares rule)


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Answer:

So {{{s^6-729}}} factors to {{{(s-3)(s+3)(s^4+9s^2+81)}}}.


In other words, {{{s^6-729=(s-3)(s+3)(s^4+9s^2+81)}}} for all values of 's'


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