Question
Given that $\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}$. Using componendo and dividendo find $a: b$.
✓
Answer
Given that $\frac{a^3+3 a b^2}{b^3+3 a^2 b}=\frac{63}{62}$
By componendo and dividendo
$
\begin{aligned}
& \frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62}=\frac{125}{1} \\
& \Rightarrow \frac{(a+b)^3}{(a b)^3}-\left(\frac{5}{1}\right)^3 \\
& \Rightarrow \frac{a+b}{a-b}=5 \\
& \Rightarrow a + b =5 a -5 b \\
& \Rightarrow 5 a - a -5 b - b =0 \\
& \Rightarrow 4 a -6 b =0 \\
& \Rightarrow 4 a =6 b \\
& \Rightarrow \frac{a}{b}=\frac{6}{4} \\
& \Rightarrow \frac{a}{b}=\frac{3}{2} \\
& a : b =3: 2 .
\end{aligned}
$
By componendo and dividendo
$
\begin{aligned}
& \frac{a^3+3 a b^2+b^3+3 a^2 b}{a^3+3 a b^2-b^3-3 a^2 b}=\frac{63+62}{63-62}=\frac{125}{1} \\
& \Rightarrow \frac{(a+b)^3}{(a b)^3}-\left(\frac{5}{1}\right)^3 \\
& \Rightarrow \frac{a+b}{a-b}=5 \\
& \Rightarrow a + b =5 a -5 b \\
& \Rightarrow 5 a - a -5 b - b =0 \\
& \Rightarrow 4 a -6 b =0 \\
& \Rightarrow 4 a =6 b \\
& \Rightarrow \frac{a}{b}=\frac{6}{4} \\
& \Rightarrow \frac{a}{b}=\frac{3}{2} \\
& a : b =3: 2 .
\end{aligned}
$
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