Answer:
Given equation:
[tex]\left(\dfrac{x^a}{x^{-b}}\right)^{a^2-ab+b^2} \times \left(\dfrac{x^b}{x^{-c}}\right)^{b^2-bc+c^2} \times \left(\dfrac{x^c}{x^{-a}}\right)^{c^2-ca+a^2}=1[/tex]
Simplify the left side of the given equation.
[tex]\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:[/tex]
[tex]\implies \left(x^{(a-(-b))}\right)^{a^2-ab+b^2} \times \left(x^{(b-(-c))}\right)^{b^2-bc+c^2} \times \left(x^{(c-(-a))}\right)^{c^2-ca+a^2}[/tex]
[tex]\implies \left(x^{(a+b)}\right)^{a^2-ab+b^2} \times \left(x^{(b+c)}\right)^{b^2-bc+c^2} \times \left(x^{(c+a)}\right)^{c^2-ca+a^2}[/tex]
[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]
[tex]\implies x^{(a+b)(a^2-ab+b^2)} \times x^{(b+c)(b^2-bc+c^2)} \times x^{(c+a)(c^2-ca+a^2)}[/tex]
Simplifying the exponents:
[tex]\begin{aligned}\implies (a+b)(a^2-ab+b^2) & =a(a^2-ab+b^2)+b(a^2-ab+b^2)\\& = a^3-a^2b+ab^2+a^2b-ab^2+b^3\\& = a^3+b^3\end{aligned}[/tex]
[tex]\begin{aligned}\implies (b+c)(b^2-bc+c^2) & =b(b^2-bc+c^2)+c(b^2-bc+c^2)\\& = b^3-b^2c+bc^2+b^2c-bc^2+c^3\\& = b^3+c^3\end{aligned}[/tex]
[tex]\begin{aligned}\implies (c+a)(c^2-ca+a^2) & =c(c^2-ca+a^2)+a(c^2-ca+a^2)\\& = c^3-c^2a+ca^2+c^2a-ca^2+a^3\\& = c^3+a^3\end{aligned}[/tex]
Therefore:
[tex]\implies x^{(a^3+b^3)} \times x^{(b^3+c^3)} \times x^{(c^3+a^3)}[/tex]
[tex]\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c}:[/tex]
[tex]\implies x^{(a^3+b^3+b^3+c^3+c^3+a^3)}[/tex]
[tex]\implies x^{(2a^3+2b^3+2c^3)}[/tex]
Therefore:
[tex]\implies x^{(2a^3+2b^3+2c^3)}[/tex]
So this cannot be proved.
