Question

If `a = x * y^(p - 1), b = x * y^(q - 1), c = x * y^(r - 1)`, prove that `a^(q - r) * b^(r - p) * c^(p - q) = 1`.

Answer 1

Given:

`a = x * y^(p - 1)`

`b = x * y^(q - 1)`

`c = x * y^(r - 1)`

To Prove:

`a^(q - r) * b^(r - p) * c^(p - q) = 1`

Proof:

Step 1: Write the expression explicitly with given values:

`a^(q - r) * b^(r - p) * c^(p - q) = (x * y^(p - 1))^(q - r) * (x * y^(q - 1))^(r - p) * (x * y^(r - 1))^(p - q)`

Step 2: Apply the power to each part inside the parentheses:

`x^(q - r) * y^((p - 1)(q - r)) * x^(r - p) * y^((q - 1)(r - p)) * x^(p - q) * y^((r - 1)(p - q))`

Step 3: Regroup powers of (x) and (y):

`x^((q - r) + (r - p) + (p - q)) * y^((p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q))`

Step 4: Simplify the exponent of (x):

(q – r) + (r – p) + (p – q) = q – r + r – p + p – q

(q – r) + (r – p) + (p – q) = 0 

So, x⁰ = 1.

Step 5: Simplify the exponent of (y):

Expand each term:

(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)

Multiply out:

pq – pr – q + r + qr – qp – r + p + rp – rq – p + q

Group and simplify terms:

• pq – qp = 0
• – pr + rp = 0
• – q + q = 0
• r – r = 0
• qr – rq = 0
• – r + r = 0
• – p + p = 0

Every term cancels, so the sum is 0.

Therefore, y⁰ = 1.

`a^(q - r) * b^(r - p) * c^(p - q) = x^0 * y^0`

`a^(q - r) * b^(r - p) * c^(p - q) = 1`

Hence proved.