Question

Prove that :
( Xa/Xb )^a2 + ab + b2 × (Xb/Xc)^b2 + bc + c2 × ( Xc/Xa )^c2 + ca + a2 = 1
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Answer 1

Consider the left hand side :

$\: \: \pink{: \implies}( \sf \dfrac{x {}^{a} }{x {}^{b}}) {}^{a {}^{2} +ab + b {}^{2}} \times( \dfrac{x {}^{b}}{x {}^{c}}) {}^{{b}^{2} + {bc}^{} + {c}^{2}} \times ( \dfrac{x {}^{2} }{x {}^{a}}) {}^{c {}^{2} + ca + a {}^{2} }$

$\pink{: \implies}\sf\dfrac{x {}^{a( {a}^{2} + ab + {b}^{2})}}{ {x}^{b( {a}^{2} + ab + {b}^{2})}} \times \dfrac{ {x}^{b( {b}^{2} + bc + {c}^{2})}}{ {x}^{c( {b}^{2} + bc + {c}^{2})}} \times \dfrac{ {x}^{c( {c}^{2} + ca + {a}^{2})}}{ {x}^{a( {c}^{2} + ca + {a}^{2})}}$

Now, Open the brackets :

$\pink{: \implies}\sf{x}^{a( {a}^{2} + ab + {b}^{2})} - {b}^{( {a}^{2} + ab + {b}^{2})} \times {x}^{b( {b}^{2} + bc + {c}^{2})} - {c}^{( {b}^{2} + bc + {c}^{2})} \times {x}^{c( {c}^{2} + ca + {a}^{2})} - {a}^{( {c}^{2} + ca + {a}^{2})}$

$\pink{:\implies}\sf{x}^{(a - b)({a}^{2} + ab + {b}^{2})} \times {x}^{(b - c)({b}^{2} + bc + {c}^{2})} \times {x}^{(c - a)({c}^{2} + ca + {a}^{2})}$

$\pink{: \implies}\sf {x}^{({a}^{3} - {b}^{3})} \times x(({{b}^{3} - c{}^{3}))} \times \:{x}^{({c}^{3} - {a}^{3})}$

$\: \: \: \: \: \: \: \: \: \pink{: \implies}\sf{x}^{({a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3})}$

Final answer :

$\: \: \boxed{ \sf\pink{: \implies}{x}^{0}} \\ \\ \: \: \: \: \: \: \: \: \: {\boxed {\boxed{{\pink{: \implies}1}}}}$

L.H.S = R.H.S

Hence, Verified.