How do you divide #(8r^3 + 27s^3) /( 4r^2 - 6rs + 9s^2)#?

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Answer 1 · 2016-04-16T06:10:47

#(8r^3+27s^3)/(4r^2-6rs+9s^2)=(2r+3s)#

To divide #(8r^3+27s^3)/(4r^2-6rs+9s^2)#, we should first factorize numerator and denominator.

As numerator is of type #a^3+b^3#, its factots will be of type #(a+b)(a^2-ab-b^2)#

Hence #(8r^3+27s^3)=[(2r)^3+(3s)^3]#

= #(2r+3s){(2r)^2-(2r)*(3s)+(3s)^2}# or

= #(2r+3s){4r^2-6rs+9s^2}#

But the latter factor is just the denominator.

Hence, #(8r^3+27s^3)/(4r^2-6rs+9s^2)#

= #((2r+3s)cancel(4r^2-6rs+9s^2))/(cancel(4r^2-6rs+9s^2))=(2r+3s)#