Question

show that;

(x^a/x^b)^a+b×(x^b/x^c)^b+c×(x^c/x^a)^c+a=1​

Answer 1

Answer:

okkk

Step-by-step explanation:

L .H .S

(Xa/x-b)a-b . (xb/x-c)b-c .(xc/x-a)c-a

= (xa+b)a-b , (xb+c)b-c .(xc+a)c-a

= x(a+b)(a-b) . x(b+c)(b-c) .x(c+a)(c-a)

= xo

= 1 = R.H.S.

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{ \bigg( \frac{ {x}^{a} }{ {x}^{b} } \bigg )^{a + b} \times \bigg( \frac{ {x}^{b} }{ {x}^{c} } \bigg)^{b + c} \times \bigg( \frac{ {x}^{c} }{ {x}^{a}} \bigg) ^{c + a} }$

$\bf \underline{To \:show-}$

$\textsf{LHS = RHS}$

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\sf{ LHS=\bigg( \frac{ {x}^{a} }{ {x}^{b} } \bigg )^{a + b} \times \bigg( \frac{ {x}^{b} }{ {x}^{c} } \bigg)^{b + c} \times \bigg( \frac{ {x}^{c} }{ {x}^{a}} \bigg) ^{c + a} }$

$\: \: \: \: \: \sf{ = \big( {x}^{a - b} \big) ^{a + b} \times \big( {x}^{b - c} \big) ^{b + c} \times \big( {x}^{c - a} \big {)}^{c + a} }$

$\: \: \: \: \: \sf{ = {x}^{ {a}^{2} - {b}^{2} } \times {x}^{ {b}^{2} - {c}^{2} } \times {x}^{ {c}^{2} - {a}^{2} } }$

$\: \: \: \: \: \sf{ = {x}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } }$

$\: \: \: \: \: \sf{ = {x}^{0} }$

$\: \: \: \: \: \sf{ = 1 =RHS }$

$\bf \underline{Hence\: proved.}$