Question

answer it............​

Answer 1

Answer:

Answer of this question ::

...................................(⌒o⌒).......................

$\huge\mathfrak{\underline{\underline{\red{Answer:-}}}}$

Step-by-step explanation:

=$\sqrt[a + b]{ \frac{x { { ^{a} }^{2} } }{x { ^{b} }^{2} } } \times \sqrt[b + c]{ \frac{ { {x}^{b} }^{2} }{ { {x}^{c} }^{2} } } \times \sqrt[c+ a]{ \frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{2} } }$

.

=$\frac{ { {x}^{a} }^{2} \frac{1}{a + b} }{ { {x}^{b} }^{2} } \times \frac{ { {x}^{b} }^{2} \frac{1}{b + c} }{ { {x}^{c} }^{2} } \times \frac{ { { {x} }^{c} }^{2} \frac{1}{c + a} }{ { { {x} }^{a} }^{2} }$

=$by \: taking \: log \: both \: sides$

=$\frac{1}{a + b} \times log \frac{ { {x}^{a} }^{2} }{ { { {x} }^{b} }^{2} } \times \frac{1}{b + c} \times log\frac{ { {x}^{b} }^{2} }{ { {x}^{c} }^{2} } \times \: \frac{1}{c + a} \times log \frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{2} }$

$\frac{1}{c + a} \times log { {x}^{a} }^{2} - log { {x}^{b} }^{2} \times \frac{1}{b + c} \times log { {x}^{b} }^{2} - log { {x}^{c} }^{2} \times \frac{1}{c + a} \times log { {x}^{c} }^{2} - log { {x}^{a} }^{2}$

$\frac{1}{c + a} \times2log {x}^{a} - 2log {x}^{b} \times \frac{1}{b + c} \times 2log {x}^{b} - 2log {x}^{c} \times \frac{1}{c + a} \times 2log {x}^{c} - 2log {x}^{a}$

=$\frac{1}{c + a} \times \: 2a( {ax}^{a -1 } ) - 2b(b {x}^{b - 1} ) + \frac{1}{b + c} + 2b(b {x}^{b - 1} ) - 2c(c {x}^{c - 1} )+ \frac{1}{c + a} \times 2c (c {x}^{c - 1} ) - 2a( {ax}^{a - 1} )$

$\frac{1}{c + a} + \frac{1}{b + c} + \frac{1}{a + b} = 0$

method (2) by differentiation:

=$\sqrt[a + b]{ \frac{ { {x}^{a} }^{2} }{ { {x}^{b} }^{2} } } \times \sqrt[b + c]{ \frac{ { {x}^{b} }^{2} }{ { {x}^{c} }^{2} } } \times \sqrt[c + a]{ \frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{?} } }$

$by \: using \: chain \: rule \:$

$\frac{dy}{dx} = \frac{ { { {x}^{a} }^{2} }^{ \frac{1}{a + b} } }{ { {x}^{b} }^{2} } \times \frac{ { { {x}^{b} }^{2} }^{ \frac{1}{b + c} } }{ { {x}^{c} }^{2} } \times \frac{ { { { {x} }^{c} }^{2} }^{ \frac{1}{c + a} } }{ { {x}^{a} }^{2} }$

$\frac{dy}{dx} = \frac{1}{a + b} \times \frac{ { {x}^{a} }^{2} }{ { {x}^{b} }^{2} } \times \frac{1}{b + c} \times \frac{ { {x}^{b} }^{2} }{ { {x}^{c} }^{2} } \times \frac{1}{c + a} \times \frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{2} }$

$\frac{dy}{dx} = \frac{1}{a + b} \times \frac{2 {x}^{a} }{2 {x}^{b} } \times \frac{1}{b + c} \times \frac{2 {x}^{b} }{2 {x}^{c} } \times \frac{1}{c + a} \times \frac{2 {x}^{c} }{ {2 {x}^{a} } }$

$\frac{dy}{dx} = \frac{1}{a + b} \times \frac{a {x}^{a - 1} }{b {x}^{b - 1} } \times \frac{1}{b + c} \times \frac{b {x}^{b - 1} }{c {x}^{c - 1} } \times \frac{1}{c + a} \times \frac{c {x}^{c - 1} }{ {ax}^{a - 1} }$

$\frac{dy}{dx} = \frac{1}{a + b} \times \frac{1}{b + c} \times \frac{1}{c + a} = 0$

$\frac{dy}{dx} = constant = 0$

method (3)

$\sqrt[a + b]{ \frac{x { {a}^{2} }}{ { {x}^{b} }^{2} } } \times \sqrt[b + c]{ \frac{ { {x}^{b} }^{2} }{ { {x}^{c} }^{2} } } \times \sqrt[c + a]{ \frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{2} } }$

$\frac{ { {x}^{a} }^{2} \frac{1}{a + b} }{ { {x}^{b} }^{2} } \times \frac{ { {x}^{b} }^{2} \frac{1}{b + c} }{ { {x}^{c} }^{2} } \times \frac{ { {x}^{c} }^{2} \frac{1}{c + a} }{ { {x}^{a} }^{2} }$

$by \: taking \: log_{e} \: both \: side$

$= \frac{1}{a + b} \times log\frac{ {x}^{ {a}^{2} } }{x {b}^{2} } \times \frac{1}{b + c} \times log \frac{ {x}^{ {b}^{2} } }{ { {x}^{c} }^{2} } \times \frac{1}{c + a} \times log\frac{ { {x}^{c} }^{2} }{ { {x}^{a} }^{2} }$

$= \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} = 0$

$= \frac{b + c + a + b}{a(b + c) + b(b + c)} + \frac{1}{c + a} = 0$

$= \frac{2b + c + a}{ab + ac + {b}^{2} + bc } + \frac{1}{c + a} = 0$

$c(2b + c + a) + ab + ac + {b}^{2} + b = 0$

$2cb + {c}^{2} + ca + ab + ac + {b}^{2} + bc = 0$

$( {c + b)}^{2} = 0$

$c = - b$

Answer 2

Sorry for the irrelevant answer but plz change ur dp...