Question

Solve the equation.
$\frac{2b^2+x^2}{b^3-x^3}-\frac{2x}{bx+b^2+x^2}+\frac{1}{x-b}=0$
For what values of b is the solution of the equation unique?

Answer 1

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

$\Rightarrow \frac{2b^2+x^2}{b^3-x^3}-\frac{2x(b-x)}{b^3-x^3}+\frac{1}{x-b}=0$

$\Rightarrow \frac{2b^2+x^2-2bx+2x^2}{b^3-x^3}=-\frac{1}{x-b}$

$\Rightarrow (2b^2-2bx+3x^2)(b-x)=(b^3-x^3)$

$\Rightarrow (b-x)(2b^2-2bx+3x^2)-(b-x)(b^2+bx+x^2)=0$

$b=x$

$\Rightarrow (2b^2-2bx+3x^2-b^2-bx-x^2)=0$

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

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

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

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

$\Rightarrow b=x$ or $b=2x$

$\therefore b=x,2x$.