If logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2) then find x^(a-b)* y^(b-c)*z^(c-a)=?

Aug 11, 2018

Let
$\log \frac{x}{{a}^{2} + a b + {b}^{2}} = \log \frac{y}{{b}^{2} + b c + {c}^{2}} = \log \frac{z}{{c}^{2} + c a + {a}^{2}} = k$

so
$\log x = k \left({a}^{2} + a b + {b}^{2}\right)$

$\log y = k \left({b}^{2} + b c + {c}^{2}\right)$

$\log z = k \left({c}^{2} + c a + {a}^{2}\right)$

Now

$\log \left[{x}^{a - b} \cdot {y}^{b - c} \cdot {z}^{c - a}\right]$

$= \left(a - b\right) \log x + \left(b - c\right) \log y + \left(c - a\right) \log z$

$= \left(a - b\right) \times k \left({a}^{2} + a b + {b}^{2}\right) + \left(b - c\right) l \times k \left({b}^{2} + b c + {c}^{2}\right) + \left(c - a\right) l \times k \left({c}^{2} + c a + {a}^{2}\right)$

$= k \left({a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3}\right) = k \times 0 = 0 = \log 1$

Hence

${x}^{a - b} \cdot {y}^{b - c} \cdot {z}^{c - a} = 1$

Aug 11, 2018

${x}^{a - b} \cdot {y}^{b - c} \cdot {z}^{c - a} = 1$

Explanation:

We know that,

$\left(1\right) {\log}_{z} A = P \iff A = {z}^{P}$

We take ,the base of log as $10$

Let ,

$\log \frac{x}{{a}^{2} + a b + {b}^{2}} = \log \frac{y}{{b}^{2} + b c + {c}^{2}} = \log \frac{z}{{c}^{2} + c a + {a}^{2}} = k$

So ,

$\log x = k \left({a}^{2} + a b + {b}^{2}\right) \implies x = {10}^{k \left({a}^{2} + a b + {b}^{2}\right)}$

logy=k(b^2+bc+c^2)=>y=10^(k(b^2+bc+c^2)

logz=k(c^2+ca+a^2)=>z=10^(k(c^2+ca+a^2)

Now ,

x^(a-b)=10^(k(a-b)(a^2+ab+b^2))=10^(k(a^3-b^3)

y^(b-c)=10^(k(b-c)(b^2+bc+c^2))=10^(k(b^3-c^3)

z^(c-a)=10^(k(c-a)(c^2+ca+a^2))=10^(k(c^3-a^3)

Hence ,

x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3) )*10^(k(b^3-c^3))*10^(k(c^3-a^3)

x^(a-b)*y^(b-c)*z^(c-a) =10^((k(a^3-b^3) )+(k(b^3-c^3))+(k(c^3-a^3)

x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3+b^3-c^3+c^3-a^3)

${x}^{a - b} \cdot {y}^{b - c} \cdot {z}^{c - a} = {10}^{k \left(0\right)} = {10}^{0} = 1$

${x}^{a - b} \cdot {y}^{b - c} \cdot {z}^{c - a} = 1$