Question

Hello people, the whole question is this, i need the third part answer ​

Answer 1

hope It will help u

Step-by-step explanation:

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Answer 2

Given equation:

`\left((x^a)/(x^(-b))\right)^(a^2-ab+b^2) * \left((x^b)/(x^(-c))\right)^(b^2-bc+c^2) * \left((x^c)/(x^(-a))\right)^(c^2-ca+a^2)=1`

Simplify the left side of the given equation.

`\textsf{Apply exponent rule} \quad (a^b)/(a^c)=a^(b-c):`

`\implies \left(x^((a-(-b)))\right)^(a^2-ab+b^2) * \left(x^((b-(-c)))\right)^(b^2-bc+c^2) * \left(x^((c-(-a)))\right)^(c^2-ca+a^2)`

`\implies \left(x^((a+b))\right)^(a^2-ab+b^2) * \left(x^((b+c))\right)^(b^2-bc+c^2) * \left(x^((c+a))\right)^(c^2-ca+a^2)`

`\textsf{Apply exponent rule} \quad (a^b)^c=a^(bc):`

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

Simplifying the exponents:

`\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}`

`\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}`

`\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}`

Therefore:

`\implies x^((a^3+b^3)) * x^((b^3+c^3)) * x^((c^3+a^3))`

`\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^(b+c):`

`\implies x^((a^3+b^3+b^3+c^3+c^3+a^3))`

`\implies x^((2a^3+2b^3+2c^3))`

Therefore:

`\implies x^((2a^3+2b^3+2c^3))`

So this cannot be proved.