If x²+y²+z² = xy+yz+zx, show that: (a^x/a^y)^{x-y}* (a^y/a^z)^{y-z} * (a^z/a^x)^{z-x} = 1 | SciPiPupil

Question: If $x²+y²+z²= xy+yz+zx$, show that: $\left ( \dfrac{a^x}{a^y} \right ) ^{x-y} ×$$\left ( \dfrac{a^y}{a^z} \right ) ^{y-z} ×$$\left ( \dfrac{a^z}{a^x} \right ) ^{z-x}$$= 1$

Solution:

Given,

$x²+y²+z²= xy+yz+zx$

To prove: $\left ( \dfrac{a^x}{a^y} \right ) ^{x-y} ×$$\left ( \dfrac{a^y}{a^z} \right ) ^{y-z} ×$$\left ( \dfrac{a^z}{a^x} \right ) ^{z-x}$$= 1$

LHS

$= \left ( \dfrac{a^x}{a^y} \right ) ^{x-y} ×\left ( \dfrac{a^y}{a^z} \right ) ^{y-z} × \left ( \dfrac{a^z}{a^x} \right ) ^{z-x}$

 

$= \left ( a^{x-y} \right ) ^{x-y}  × \left ( a^{y-*} \right ) ^{y-z} ×  \left ( a^{z-x} \right ) ^{z-x}$

$= a^{(x-y)(x-y)} × a^{(y-z)(y-z)} × a^{(z-x)(z-x)}$

$= a^{(x-y)² + (y-z)² + (z-x)²}$

$= a^{x² -2xy+y²+y²-2yz +z²+z²-2zx+x²}$

$= a^{2(x²+y²+z²) - 2(xy+yz+zx)}$

[ Given: x²+y²+z² = xy+yz+zx ]

$= a^{2(xy+yz+zx) - 2(xy+yz+zx)}$

$= a^0$

$= 1$

RHS

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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