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

Question

1479 views around the world

You can reuse this answer
Creative Commons License

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

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

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