Question

If $a=x^{m+n}\cdot x^l;b=x^{n+l}\cdot x^m$ and $c=x^{l+m}\cdot x^n$. Prove that: $a^{m-n}\cdot b^{n-l}\cdot c^{l-m}=1$

Answer 1

Step : Simplifying the LHS of given expression
Given,
$a=x^{m+n}\cdot x^l;b=x^{n+l}\cdot x^m$ and $c=x^{l+m}\cdot x^n$
To prove : $a^{m-n}\cdot b^{n-l}\cdot c^{l-m}=1$.
Taking L.H.S, we have:
$\begin{array}{cc}& =a^{m-n}\cdot b^{n-l}\cdot c^{l-m}\\ & ={(x^{m+n}\cdot x^l)}^{m-n}\cdot {(x^{n+l}\cdot x^m)}^{n-l}\cdot {(x^{l+m}\cdot x^n)}^{l-m}\\ & =(x^{(m+n)(m-n)}\cdot x^{l(m-n)})\cdot (x^{(n+l)(n-l)}\cdot x^{m(n-l)}),\\ & (x^{(l+m)(l-m)}\cdot x^{n(l-m)})\\ & =(x^{(m^2-n^2)}\cdot x^{lm-ln})\cdot (x^{(n^2-l^2)}\cdot x^{nm-lm}).\\ & (x^{(l^2-m^2)}\cdot x^{nl-nm})\\ & =x^{(m^2-n^2+lm-ln+n^2-l^2+nm-lm+l^2-m^2+nl-nm)}\\ & =x^0=1=\text{R.H.S}\end{array}$
Hence proved.