Question 147759

{{{(2z+5)(4z^2-10z+25)}}} Start with the given expression.



{{{2z(4z^2-10z+25)+5(4z^2-10z+25)}}} Expand.



{{{(2z)*(4z^2)+(2z)*(-10z)+(2z)*(25)+(5)*(4z^2)+(5)*(-10z)+(5)*(25)}}} Distribute.



{{{8*z^3-20*z^2+50*z+20*z^2-50*z+125}}} Multiply.



{{{8*z^3+125}}} Now combine like terms.


 
Notice how {{{8*z^3+125}}} is a binomial where both terms are cubes of other terms. In other words, {{{8*z^3=(2z)^3}}} and {{{125=(5)^3}}}. The short cut you mentioned would use the sum of cubes formula {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}} to expand the original problem (which is a much faster way to solve the problem).



So {{{(2z+5)(4z^2-10z+25)}}} expands to {{{8*z^3+125}}}.
 
 


In other words, {{{(2z+5)(4z^2-10z+25)=8*z^3+125}}}.