If $a,b,c,d$ are continued proportion : Prove that : $(\frac{a-b}{c}+\frac{a-c}{b})^2-(\frac{d-b}{c}+\frac{d-c}{b})^2=(a-d)(\frac{1}{c^2}-\frac{1}{b^2})^2$
After solving LH.S I got : $\frac{2(a-d)}{(bc)^2}$
But after solving R.H.S I am getting $\frac{(a-d)^2(b^2-c^2)}{(bc)^2}$
Please help.....
May be this term : $(a-d)(\frac{1}{c^2}-\frac{1}{b^2})^2$ is $(a-d)(\frac{1}{c^2}-\frac{1}{b^2})$ not square of term
If $a,b,c,d$ are in continued proportion, we can divide through by $a$ and make them $1,r,r^2,r^3$. Then: $$\begin {align} \left(\frac {a-b}c+\frac{a-c}b\right)^2-\left(\frac{d-b}c+\frac {d-c}b\right)^2&= \left(\frac {1-r}{r^2}+\frac{1-r^2}r\right)^2-\left( \frac {r^3-r}{r^2}+\frac{r^3-r^2}r\right)^2 \\&=\left( \frac{1-r^3}{r^2}\right)^2-\left(\frac{r^4-r}{r^2}\right)^2\\&=\frac{(1+r)^2(1-r^3)^2}{r^4}\\&=\frac{(r+r^2)^2(1-r^3)^2}{r^6}\\&=\frac{(b+c)^2(a-d)^2}{b^2c^2}\end {align}$$
Which differs from what you got for the right by a sign.