Question

Solve the equation $\frac{2b^2+x^2}{b^3-x^3}-\frac{2x}{bx+b^2+x^2}+\frac{1}{x-b}=0$

Answer 1

$\frac{2b^2+x^2}{b^3-x^3}-\frac{2x}{x^2+b^2+bx}+\frac{1}{x-b}=0$ is defined for $x\ne b$

$\Rightarrow \frac{2b^2+x^2}{-(x-b)(x^2+b^2+bx)}-\frac{2x}{x^2+b^2+bx}+\frac{1}{x-b}=0$ and $x\ne b$

$\Rightarrow \frac{-2b^2-x^2}{(x-b)(x^2+b^2+bx)}-\frac{2x}{x^2+b^2+bx}+\frac{1}{x-b}=0$ and $x\ne b$

$\Rightarrow \frac{-2b^2-x^2-2x(x-b)+x^2+b^2+bx}{(x-b)(x^2+b^2+bx)}=0$ and $x\ne b$

$\Rightarrow -2b^2-x^2-2x(x-b)+x^2+b^2+bx=0,$ and $x\ne b$

$\Rightarrow -2b^2-x^2-2x^2+2xb+x^2+b^2+bx=0$

$\Rightarrow -b^2-2x^2+3xb=0$

$\Rightarrow 2x^2-3xb+b^2=0$

$\Rightarrow 2x^2-2xb-xb+b^2=0$

$\Rightarrow 2x(x-b)-b(x-b)=0$

$\Rightarrow (x-b)(2x-b)=0$

$\Rightarrow x-b=0,2x=b$

$\therefore x=b,\frac{b}{2}$

But $x\ne b$

$\therefore x=\frac{b}{2}$