Question 144503
{{{(2x+3)^3 - 64}}} Start with the given expression.



Let a=2x+3



So we now have {{{a^3 - 64 }}}



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



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



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




{{{(2x+3-4)((2x+3)^2+4(2x+3)+16)}}} Now replace each "a" with {{{2x+3}}}



{{{(2x+3-4)(4x^2+12x+9+4(2x+3)+16)}}} Foil



{{{(2x+3-4)(4x^2+12x+9+8x+12+16)}}} Distribute



{{{(2x-1)(4x^2+20x+37)}}} Combine like terms


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Answer:

So {{{(2x+3)^3 - 64}}} factors to {{{(2x-1)(4x^2+20x+37)}}}.


In other words, {{{(2x+3)^3 - 64=(2x-1)(4x^2+20x+37)}}}