Question:

$\left(\frac{x+1}{x-4}\right)^{2}=\frac{x+8}{x-2}$

Solution:

$\left(\frac{x+1}{x-4}\right)^{2}=\frac{x+8}{x-2}$

or $\frac{\mathrm{x}^{2}+2 \mathrm{x}+1}{\mathrm{x}^{2}-8 \mathrm{x}+16}=\frac{\mathrm{x}+8}{\mathrm{x}-2} \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right.$ and $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

or $\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}-2 \mathrm{x}^{2}-4 \mathrm{x}-2=\mathrm{x}^{3}-8 \mathrm{x}^{2}+16 \mathrm{x}+8 \mathrm{x}^{2}-64 \mathrm{x}+128$ [After cross multiplication]

or $\mathrm{x}^{3}-\mathrm{x}^{3}-3 \mathrm{x}+48 \mathrm{x}=128+2$

or $45 \mathrm{x}=130$

or $\mathrm{x}=\frac{130}{45}=\frac{26}{9}$

Thus $\mathrm{x}=\frac{26}{9}$ is the solution of the given equation.

Check :

Substituting $\mathrm{x}=\frac{26}{9}$ in the given equation, we get :

L.H. S. $=\left(\frac{\frac{26}{9}+1}{\frac{26}{9}-4}\right)^{2}=\left(\frac{26+9}{26-36}\right)^{2}=\frac{1225}{100}=\frac{49}{4}$

R.H.S. $=\left(\frac{\frac{26}{9}+8}{\frac{26}{9}-2}\right)=\left(\frac{26+72}{26-18}\right)=\frac{98}{8}=\frac{49}{4}$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{x}=\frac{26}{9}$