Question 168970

{{{b^6-1}}} Start with the given expression.



{{{(b^2)^3-(1)^3}}} Rewrite {{{b^6}}} as {{{(b^2)^3}}}. Rewrite {{{1}}} as {{{(1)^3}}}.



{{{(b^2-1)((b^2)^2+(b^2)(1)+(1)^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)}}}



{{{(b^2-1)(b^4+b^2+1)}}} Multiply



{{{(b+1)(b-1)(b^4+b^2+1)}}} Factor {{{b^2-1}}} to get {{{(b+1)(b-1)}}} (by use of the difference of squares)



{{{(b+1)(b-1)(b^2+b+1)(b^2-b+1)}}} Factor {{{b^4+b^2+1}}} to get {{{(b^2+b+1)(b^2-b+1)}}} 


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Answer:


So {{{b^6-1}}} completely factors to {{{(b+1)(b-1)(b^2+b+1)(b^2-b+1)}}}.


In other words, {{{b^6-1=(b+1)(b-1)(b^2+b+1)(b^2-b+1)}}}