Question

Solve this algebra...​

Answer 1

Explanation:

`\text{Use}\ (a^n)/(a^m)=a^(n-m)\ \text{and}\ (a^n)^m=a^(nm)\\\\\left((a^x)/(a^y)\right)^(x-y)=\left(a^(x-y)\right)^(x-y)=a^((x-y)(x-y))=a^((x-y)^2)\\\\\left((a^y)/(a^z)\right)^(y-z)=\left(a^(y-z)\right)^(y-z)=a^((y-z)(y-z))=a^((y-z)^2)\\\\\left((a^z)/(a^x)\right)^(z-x)=\left(a^(z-x)\right)^(z-x)=a^((z-x)(z-x))=a^((z-x)^2)\\\\\text{Use}\ a^n\cdot a^m=a^(n+m)\\\\a^((x-y)^2)\cdot a^((y-z)^2)\cdot a^((z-x)^2)=a^((x-y)^2+(y-z)^2+(z-x)^2)`

`\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-y)^2=x^2-2xy+y^2\\(y-z)^2=y^2-2yz+z^2\\(z-x)^2=z^2-2zx+x^2\\\\=a^(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2)\\\\\text{combine like terms}\\\\=a^((x^2+x^2)+(y^2+y^2)+(z^2+z^2)-2xy-2yz-2zx)\\\\=a^(2x^2+2y^2+2z^2-2xy-2yz-2zx)\qquad\text{distributive}\\\\=a^(2(x^2+y^2+z^2)-2(xy-yz-zx))\\\\\text{From the equastion we know:}\ x^2+y^2+z^2=xy+yx+zx.\\\text{Therefore}\\\\=a^(2(x^2+y^2+z^2)-2(x^2+y^2+z^2))=a^0=1`