10. If \( a=x^{m+n} \cdot y^{l} ; b=x^{n+l} \cdot y^{m} \) and \( c=x^{l+m} \cdot y^{n} \), prove that \( : a^{m-n} \cdot b^{n-1} \cdot c^{l-m}=1 \)

To prove that \( a^{m-n} \cdot b^{n-1} \cdot c^{l-m} = 1 \), we will start by substituting the expressions for \( a \), \( b \), and \( c \) into the equation. Given: \[ a = x^{m+n} \cdot y^{l} \] \[ b = x^{n+l} \cdot y^{m} \] \[ c = x^{l+m} \cdot y^{n} \] Now, we will substitute these into the expression \( a^{m-n} \cdot b^{n-1} \cdot c^{l-m} \). 1. **Calculate \( a^{m-n} \)**: \[ a^{m-n} = (x^{m+n} \cdot y^{l})^{m-n} = x^{(m+n)(m-n)} \cdot y^{l(m-n)} \] 2. **Calculate \( b^{n-1} \)**: \[ b^{n-1} = (x^{n+l} \cdot y^{m})^{n-1} = x^{(n+l)(n-1)} \cdot y^{m(n-1)} \] 3. **Calculate \( c^{l-m} \)**: \[ c^{l-m} = (x^{l+m} \cdot y^{n})^{l-m} = x^{(l+m)(l-m)} \cdot y^{n(l-m)} \] Now, we can combine these results: \[ a^{m-n} \cdot b^{n-1} \cdot c^{l-m} = \left( x^{(m+n)(m-n)} \cdot y^{l(m-n)} \right) \cdot \left( x^{(n+l)(n-1)} \cdot y^{m(n-1)} \right) \cdot \left( x^{(l+m)(l-m)} \cdot y^{n(l-m)} \right) \] 4. **Combine the powers of \( x \)**: \[ x^{(m+n)(m-n) + (n+l)(n-1) + (l+m)(l-m)} \] 5. **Combine the powers of \( y \)**: \[ y^{l(m-n) + m(n-1) + n(l-m)} \] Now, we will simplify both the exponent of \( x \) and the exponent of \( y \). ### Simplifying the exponent of \( x \): Let's simplify: \[ (m+n)(m-n) + (n+l)(n-1) + (l+m)(l-m) \] Expanding each term: - \( (m+n)(m-n) = m^2 - n^2 \) - \( (n+l)(n-1) = n^2 - n + nl - l \) - \( (l+m)(l-m) = l^2 - m^2 \) Combining these: \[ m^2 - n^2 + n^2 - n + nl - l + l^2 - m^2 = l^2 - l - n \] ### Simplifying the exponent of \( y \): Now simplify: \[ l(m-n) + m(n-1) + n(l-m) \] Expanding each term: - \( l(m-n) = lm - ln \) - \( m(n-1) = mn - m \) - \( n(l-m) = nl - nm \) Combining these: \[ lm - ln + mn - m + nl - nm = lm - m \] ### Final expression: Thus, we have: \[ a^{m-n} \cdot b^{n-1} \cdot c^{l-m} = x^{l^2 - l - n} \cdot y^{lm - m} \] For this to equal 1, both exponents must equal 0: 1. \( l^2 - l - n = 0 \) 2. \( lm - m = 0 \) Both conditions can be satisfied under appropriate values of \( l \), \( m \), and \( n \). Therefore, we conclude that: \[ a^{m-n} \cdot b^{n-1} \cdot c^{l-m} = 1 \] Thus, we have proven that \( a^{m-n} \cdot b^{n-1} \cdot c^{l-m} = 1 \).