Question

$\frac{ 1}{1 + x {}^{b - a} + x {}^{c - a} } + \frac{1}{1 + x {}^{a -b } + x {}^{c - b} } + \frac{1}{1 + x {}^{b - c} + x {}^{a - c} } = 1$
prove the statement with proper explanation plzzzz.....

Answer 1

Hello Mate!

$\frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c} } = 1 \\ \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{a} } + \frac{ {x}^{c} }{ {x}^{a} } } + \frac{1}{1 + \frac{ {x}^{a } }{ {x}^{b} } + \frac{ {x}^{c} }{ {x}^{b} } } + \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{c} } + \frac{ {x}^{a} }{ {x}^{c} } } = 1 \\ \frac{ {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{ {x}^{b} }{ {x}^{b} + {x}^{a} + {x}^{c} } + \frac{ {x}^{c} }{ {x}^{c} + {x}^{b} + {x}^{a} } = 1 \\ \frac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} } = 1 \\ 1 = 1$

Hope it helps☺!

Answer 2

Step-by-step explanation:

Prove that:

$\frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{a - c} + {x}^{b - c} } = 1$

We have,

$\frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{a - c} + {x}^{b - c} } = 1$

$⟹ \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{a} } + \frac{ {x}^{c} }{ {x}^{a} } } + \frac{1}{1 + \frac{ {x}^{a} }{ {x}^{b} } + \frac{ {x}^{c} }{ {x}^{b} } } + \frac{1}{1 + \frac{ {x}^{a} }{ {x}^{c} } + \frac{ {x}^{b} }{ {x}^{c} } } = 1$

$⟹ \frac{ {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{ {x}^{b} }{ {x}^{b} + {x}^{a} + {x}^{c} } + \frac{ {x}^{c} }{ {x}^{c} + {x}^{a} + {x}^{b} } = 1$

$⟹ \frac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} } = 1$

$⟹ 1 = 1$

LHS = RHS

Hence, proved: