Question

Solve ${\displaystyle \frac{1+x+x^2}{1-x+x^2}=\frac{62(1+x)}{63(1+x)}}$

Answer 1

${\displaystyle \frac{1+x+x^2}{1-x+x^2}=\frac{62(1+x)}{63(1+x)}}$

⇒ ${\displaystyle \frac{(1-x)(1+x+x^2)}{(1+x)(1-x+x^2)}=\frac{62}{63}}$

⇒ ${\displaystyle \frac{(1+x)(1-x+x^2)}{(1-x)(1+x+x^2)}=\frac{62}{63}}$

⇒ ${\displaystyle \frac{1+x^3}{1-x^3}=\frac{63}{62}}$
Applying componendo and dividendo,

${\displaystyle \frac{1+x^3+1-x^3}{1+x^3-1+x^3}=\frac{63+62}{63-62}}$

⇒ ${\displaystyle \frac{2}{2x^3}=\frac{125}{1}}$

⇒ ${\displaystyle \frac{1}{x^3}=\frac{125}{1}}$

⇒ x³ = ${\displaystyle \frac{1}{125}}$

= ${\displaystyle {\left(\frac{1}{5}\right)}^3}$

∴ x = ${\displaystyle \frac{1}{5}}$.