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

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

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Answer 1 · 2016-04-16T06:10:47

$\frac{8 r^{3} + 27 s^{3}}{4 r^{2} - 6 r s + 9 s^{2}} = (2 r + 3 s)$ $(8r^3+27s^3)/(4r^2-6rs+9s^2)=(2r+3s)$

Explanation:

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

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

Hence $(8 r^{3} + 27 s^{3}) = [{(2 r)}^{3} + {(3 s)}^{3}]$ $(8r^3+27s^3)=[(2r)^3+(3s)^3]$

= $(2 r + 3 s) {{(2 r)}^{2} - (2 r) \cdot (3 s) + {(3 s)}^{2}}$ $(2r+3s){(2r)^2-(2r)*(3s)+(3s)^2}$ or

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

But the latter factor is just the denominator.

Hence, $\frac{8 r^{3} + 27 s^{3}}{4 r^{2} - 6 r s + 9 s^{2}}$ $(8r^3+27s^3)/(4r^2-6rs+9s^2)$

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

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

1 Answer

Explanation:

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Explanation:

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