Question

Prove the following:
(c+a-b)
ctb-a
bta-c
xa
1
(
x6
(i)
xo
xb
ro
xo
xa​

Answer 1

$\star\:\:\:\bf\large\underline\blue{Given\:To\:Prove:-}$

• $\small\small {( \frac{ {x}^{b} }{ {x}^{a} } })^{b + a - c} \times ( \frac{ {x}^{c} }{ {x}^{b} } )^{b + c - a} \times {( \frac{ {x}^{a} }{ {x}^{c} } )}^{c + a - b} = 1$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

$\star\:\:\:\bf\underline\blue{LHS:-}$

$\small\small {( \frac{ {x}^{b} }{ {x}^{a} } })^{b + a - c} \times ( \frac{ {x}^{c} }{ {x}^{b} } )^{b + c - a} \times {( \frac{ {x}^{a} }{ {x}^{c} } )}^{c + a - b}$

$\small\small {( {x}^{b - a} })^{b + a - c} \times ( {x}^{c - b})^{b + c - a} \times {( {x}^{a - c})}^{c + a - b}$

$\small= {x}^{ {b}^{2} - {a}^{2} + ac - bc } \times {x}^{ {c}^{2} - {b}^{2} + ab - ac} \times {x}^{ {a}^{2} - {c}^{2} + bc - ab}$

$\small= {x}^{ \cancel {{b}^{2} - {a}^{2} + ac - bc + {c}^{2} - {b}^{2} + ab - ac + {a}^{2} - {c}^{2} + bc - ab }}$

$= {x}^{0}$

$= 1$

$\star\:\:\:\bf\underline\blue{RHS:-}$

$= 1$

Therefore,

$\small\small {( \frac{ {x}^{b} }{ {x}^{a} } })^{b + a - c} \times ( \frac{ {x}^{c} }{ {x}^{b} } )^{b + c - a} \times {( \frac{ {x}^{a} }{ {x}^{c} } )}^{c + a - b} = 1$

Hence Proved.

Answer 2

Answer:

hey hope u got ur answer afte ethos pic