Question

Prove that `(x^a/x^b)^(a^2 + ab + b^2) (x^b/x^c)^(b^2 + bc + c^2) (x^c/x^a)^(c^2 + ca + a^2) = 1`.

Answer 1

Given: `(x^a/x^b)^(a^2 + ab + b^2) (x^b/x^c)^(b^2 + bc + c^2) (x^c/x^a)^(c^2 + ca + a^2) = ?`

To Prove: `(x^a/x^b)^(a^2 + ab + b^2) (x^b/x^c)^(b^2 + bc + c^2) (x^c/x^a)^(c^2 + ca + a^2) = 1`

Proof (Stepwise):

1. Express each term using exponent laws:

`(x^a/x^b)^(a^2 + ab + b^2) = (x^(a - b))^(a^2 + ab + b^2)`

`(x^a/x^b)^(a^2 + ab + b^2) = x^((a - b)(a^2 + ab + b^2)`

`(x^b/x^c)^(b^2 + bc + c^2) = (x^(b - c))^(b^2 + bc + c^2)`

`(x^b/x^c)^(b^2 + bc + c^2) = x^((b - c)(b^2 + bc + c^2)`

`(x^c/x^a)^(c^2 + ca + a^2) = (x^(c - a))^(c^2 + ca + a^2)`

`(x^c/x^a)^(c^2 + ca + a^2) = x^((c - a)(c^2 + ca + a^2))`

2. Combine all three to get the total exponent of (x):

`x^((a - b)(a^2 + ab + b^2) + (b - c)(b^2 + bc + c^2) + (c - a)(c^2 + ca + a^2)`

3. Let us denote the total exponent as:

E = (a – b)(a² + ab + b²) + (b – c)(b² + bc + c²) + (c – a)(c² + ca + a²)

4. Expand each term of (E):

(a – b)(a² + ab + b²) = a³ + a²b + ab² – ba² – b²a – b³

(b – c)(b² + bc + c²) = b³ + b²c + bc² – cb² – c²b – c³

(c – a)(c² + ca + a²) = c³ + c²a + ca² – ac² – a²c – a³

5. Add the expansions:

E = (a³ + a²b + ab² – ba² – b²a – b³) + (b³ + b²c + bc² – cb² – c²b – c³) + (c³ + c²a + ca² – ac² – a²c – a³)

6. Rearranging terms, group like terms to see cancellations:

Notice (a³) and (–a³), (b³) and (–b³), (c³) and (–c³) cancel out.

Group the terms with (a²b, ab², b²c, bc², c²a, ca²) and symmetrical negative terms.

Rewrite:

E = (a²b + ab² – b²a – ba²) + (b²c + bc² – cb² – c²b) + (c²a + ca² – ac² – a²c)

7. Each grouped pair cancels as follows:

a²b – ba² = 0, since a²b = ba²

ab² – b²a = 0

Similarly for other groups:

b²c – cb² = 0

bc² – c²b = 0

c²a – ac² = 0

ca² – a²c = 0

8. Therefore, total exponent E = 0.

Hence, x^E = x⁰ = 1.