Question

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

• A. $8:3$
• B. $9:1$
• C. $11:1$
• D. $7:2$

Answer 1

Answer: (B)

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