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  4. If (p3+3 p q2/3 p2 q+q3)=(189/61), then find the value of p: q

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Q. If $\frac{p^3+3 p q^2}{3 p^2 q+q^3}=\frac{189}{61}$, then find the value of $p: q$

Ratio, Proportion and Variation

Solution:

$\frac{p^3+3 p q^2}{3 p^2 q+q^3}=\frac{189}{61}$
On applying componendo and dividendo
$\frac{p^3+3 p q^2+3 p^2 q+q^3}{p^3+3 q^2-3 p^2 q-q^3}=\frac{189+61}{189-61} $
$\Rightarrow \frac{(p+q)^3}{(p-q)^3}=\frac{250}{128} \Rightarrow \frac{(p+q)^3}{(p-q)^3}=\frac{125}{64}$
$\Rightarrow \frac{p+q}{p-q}=\frac{5}{4}$
On applying componendo and dividendo
$\Rightarrow \frac{p+q+p-q}{p+q-p+q}=\frac{5+4}{5-4}$
$\therefore p: q=9: 1 .$