Show that: (a+1 b )^m (a-1 b )^n (b+1 a )^m (b-1 a )^n = ( a b )^ m+n

LHS= (a+1 b )^m (a-1 b )^n (b+1 a )^m (b-1 a )^n = ( ab+1 b )^m ( ab-1 b )^n ( ab+1 a )^m ( ab-1 a )^ n = (ab+1)^m b^m (ab-1)^n b^n (ab+1)^m a^m (ab-1)^n a^n = (ab+1)^m b^m (ab-1)^n b^n a^m (ab+1)^m a^n (ab-1)^n = a^ m+n b^ m+n = ( a b )^ m+n =RHS

Show that: (a+1 b )^m (a-1 b )^n (b+1 a )^m (b-1 a )^n = ( a b )^…

Download App

Question

Show that:$\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$

✓

Answer

$\text{LHS}=\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}$$=\frac{\Big(\frac{\text{ab}+1}{\text{b}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{b}}\Big)^\text{n}}{\Big(\frac{\text{ab}+1}{\text{a}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{a}}\Big)^{\text{n}}}$
$=\frac{\frac{\text{(ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}}{\frac{(\text{ab}+1)^\text{m}}{\text{a}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{a}^\text{n}}}$
$=\frac{(\text{ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}\times\frac{\text{a}^\text{m}}{(\text{ab}+1)^\text{m}}\times\frac{\text{a}^\text{n}}{(\text{ab}-1)^\text{n}}$
$=\frac{\text{a}^{\text{m}+\text{n}}}{\text{b}^{\text{m}+\text{n}}}$
$=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$
$=\text{RHS}$