Given that [tex]\( x^3 + y^3 + z^3 = 1 \)[/tex], we need to prove that:
[tex]\[
\left( \frac{a^x}{a^{-y}} \right)^{x^2 - xy + y^2} \times \left( \frac{a^y}{a^{-z}} \right)^{y^2 - yz + z^2} \times \left( \frac{a^z}{a^{-x}} \right)^{z^2 - zx + x^2} = a^2
\][/tex]
### Simplify Each Term
1. Simplify the bases:
[tex]\[
\frac{a^x}{a^{-y}} = a^{x - (-y)} = a^{x + y}
\][/tex]
[tex]\[
\frac{a^y}{a^{-z}} = a^{y - (-z)} = a^{y + z}
\][/tex]
[tex]\[
\frac{a^z}{a^{-x}} = a^{z - (-x)} = a^{z + x}
\][/tex]
2. Rewrite the equation with simplified bases:
[tex]\[
\left( a^{x + y} \right)^{x^2 - xy + y^2} \times \left( a^{y + z} \right)^{y^2 - yz + z^2} \times \left( a^{z + x} \right)^{z^2 - zx + x^2}
\][/tex]
3. Combine the exponents:
Recall that [tex]\((a^m)^n = a^{mn}\)[/tex]:
[tex]\[
a^{(x+y)(x^2 - xy + y^2)} \times a^{(y+z)(y^2 - yz + z^2)} \times a^{(z+x)(z^2 - zx + x^2)}
\][/tex]
4. Add the exponents:
Since the bases are the same ([tex]\( a \)[/tex]), we can combine the exponents:
[tex]\[
a^{(x+y)(x^2 - xy + y^2) + (y+z)(y^2 - yz + z^2) + (z+x)(z^2 - zx + x^2)}
\][/tex]
### Expand and Simplify the Exponents
Next, we'll expand the expressions inside the exponents:
[tex]\[
(x+y)(x^2 - xy + y^2)
\][/tex]
Expanding, we get:
[tex]\[
x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3 = x^3 + y^3
\][/tex]
Similarly for:
[tex]\[
(y+z)(y^2 - yz + z^2)
\][/tex]
Expanding, we get:
[tex]\[
y^3 - y^2z + yz^2 + y^2z - yz^2 + z^3 = y^3 + z^3
\][/tex]
And:
[tex]\[
(z+x)(z^2 - zx + x^2)
\][/tex]
Expanding, we get:
[tex]\[
z^3 - z^2x + zx^2 + z^2x - zx^2 + x^3 = z^3 + x^3
\][/tex]
### Combine All Expanded Terms
Now, add all the expanded terms:
[tex]\[
x^3 + y^3 + y^3 + z^3 + z^3 + x^3 = 2(x^3 + y^3 + z^3)
\][/tex]
Given [tex]\( x^3 + y^3 + z^3 = 1 \)[/tex], substitute:
[tex]\[
2(x^3 + y^3 + z^3) = 2 \cdot 1 = 2
\][/tex]
Thus, the combined exponent simplifies to [tex]\(2\)[/tex], yielding:
[tex]\[
a^{2}
\][/tex]
### Conclusion
Therefore, we have proven that:
[tex]\[
\left( \frac{a^x}{a^{-y}} \right)^{x^2 - xy + y^2} \times \left( \frac{a^y}{a^{-z}} \right)^{y^2 - yz + z^2} \times \left( \frac{a^z}{a^{-x}} \right)^{z^2 - zx + x^2} = a^2
\][/tex]
This concludes the proof.
