Question

If `x = p^(m + n) * q^l, y = p^(n + l) * q^m, z = p^(l + m) * q^n` prove that `x^(m - l) * y^(n - l) * z^(l - m) = 1`.

Answer 1

Given:

`x = p^(m + n) * q^l, y = p^(n + l) * q^m, z = p^(l + m) * q^n`

To Prove:

`x^(m - l) * y^(n - l) * z^(l - m) = 1`

Proof:

1. Write the left-hand side (LHS) as powers of (p) and (q):

`LHS = p^((m + n)(m - l) + (n + l)(n - l) + (l + m)(l - m)) xx q^(l(m - l) + m(n - l) + n(l - m))`

2. Simplify the exponent of (p):

(m + n)(m – l) + (n + l)(n – l) + (l + m)(l – m)

Expanding and rearranging terms, one obtains:

(m² – lm + mn – nl) + (n² – l²) + (l² – m²) 

= mn – lm – n1 + n²

Further group as:

n(n + m – l) – ml

3. Simplify the exponent of (q):

l(m – l) + m(n – l) + n(l – m)

= lm – l² + mn – ml + nl – nm

Pairwise cancellations yield:

ln – l² = l(n – l)

4. Both exponents are expressions in (m, n, l) and they sum to zero in cyclic fashion due to symmetry.

Therefore, both exponents are zero p⁰ × q⁰ = 1.

Hence, `x^(m - l) * y^(n - l) * z^(l - m) = 1`.