Question

Given that $\frac{a^3+3{ab}^2}{b^3+3a^{2}b}=\frac{63}{62}$. Using componendo and dividendo find $a:b$ and write the value of $\frac{a}{b}+\frac{1}{2}$

Answer 1

$\frac{a^3+3{ab}^2}{b^3+3a^{2}b}=\frac{63}{62}$

Using Componendo and DIvidendo

$\Rightarrow \frac{(a^3+3{ab}^2)+(b^3+3a^{2}b)}{(a^3+3{ab}^2)-(b^3+3a^{2}b)}=\frac{63+62}{63-62}$

$\Rightarrow \frac{{(a+b)}^3}{{(a-b)}^3}=125\Rightarrow {\left(\frac{a+b}{a-b}\right)}^3=5^3$

$\Rightarrow \frac{a+b}{a-b}=5\Rightarrow a+b=5(a-b)\Rightarrow 4a=6b\Rightarrow \frac{a}{b}=\frac{6}{4}\therefore a:b=3:2$