Question 167704
Let {{{z=b-2}}} 


So the expression {{{64 - (b-2)^3}}} then becomes {{{64-z^3}}}



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{{{64-z^3}}} Start with the given expression.



{{{(4)^3-(z)^3}}} Rewrite {{{64}}} as {{{(4)^3}}}. Rewrite {{{z^3}}} as {{{(z)^3}}}.



{{{(4-z)((4)^2+(4)(z)+(z)^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)}}}



Note: in this case, {{{A=4}}} and {{{B=z}}}




{{{(4-z)(16+4z+z^2)}}} Multiply



{{{(4-(b-2))(16+4(b-2)+(b-2)^2)}}} Plug in {{{z=b-2}}} 



{{{(4-(b-2))(16+4(b-2)+b^2-4b+4)}}} FOIL



{{{(4-b+2)(16+4b-8+b^2-4b+4)}}} Distribute



{{{(-b+6)(b^2+12)}}} Combine like terms.



{{{-(b-6)(b^2+12)}}} Factor a negative 1 from {{{-b+6}}} to get {{{-(b-6)}}}




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Answer:



So {{{64 - (b-2)^3}}} factors to {{{-(b-6)(b^2+12)}}}