The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?\eqalign{ & {\text{The value of}} \cr & {\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}{\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.{\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}} \cr} " /> The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?\eqalign{ & {\text{The value of}} \cr & {\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}{\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.{\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}} \cr} " /> The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?The value of(xb+cc−a)1a−b.(xc+aa−b)1b−c.(xa+bb−c)1c−a is = ?\eqalign{ & {\text{The value of}} \cr & {\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}{\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.{\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}} \cr} " />

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$\begin{aligned} The value of \\ {(x^{\frac{b + c}{c - a}})}^{\frac{1}{a - b}} . {(x^{\frac{c + a}{a - b}})}^{\frac{1}{b - c}} . {(x^{\frac{a + b}{b - c}})}^{\frac{1}{c - a}} is = ? \end{aligned}$ $\eqalign{ & {\text{The value of}} \cr & {\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}{\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.{\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}} \cr}$

11
2a
3b
4c

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{\frac{b + c}{(a - b) (c - a)}} . x^{\frac{c + a}{(a - b) (b - c)}} . x^{\frac{a + b}{(b - c) (c - a)}} \\ = x^{\frac{(b + c) (b - c) + (c + a) (c - a) + (a + b) (a - b)}{(a - b) (b - c) (c - a)}} \\ = x^{\frac{(b^{2} - c^{2}) + (c^{2} - a^{2}) + (a^{2} - b^{2})}{(a - b) (b - c) (c - a)}} \\ = x^{0} \\ = 1 \end{aligned}

$\eqalign{ & {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^0} \cr & = 1 \cr}$

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", "text": "

\begin{aligned} The value of \\ {(x^{\frac{b + c}{c - a}})}^{\frac{1}{a - b}} . {(x^{\frac{c + a}{a - b}})}^{\frac{1}{b - c}} . {(x^{\frac{a + b}{b - c}})}^{\frac{1}{c - a}} is  =  ? \end{aligned}

$\eqalign{ & {\text{The value of}} \cr & {\left( {{x^{\frac{{b + c}}{{c - a}}}}} \right)^{\frac{1}{{a - b}}}}{\text{.}}{\left( {{x^{\frac{{c + a}}{{a - b}}}}} \right)^{\frac{1}{{b - c}}}}.{\left( {{x^{\frac{{a + b}}{{b - c}}}}} \right)^{\frac{1}{{c - a}}}}{\text{ is = ?}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x^{\frac{b + c}{(a - b) (c - a)}} . x^{\frac{c + a}{(a - b) (b - c)}} . x^{\frac{a + b}{(b - c) (c - a)}} \\ = x^{\frac{(b + c) (b - c) + (c + a) (c - a) + (a + b) (a - b)}{(a - b) (b - c) (c - a)}} \\ = x^{\frac{(b^{2} - c^{2}) + (c^{2} - a^{2}) + (a^{2} - b^{2})}{(a - b) (b - c) (c - a)}} \\ = x^{0} \\ = 1 \end{aligned}

$\eqalign{ & {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^0} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x^{\frac{b + c}{(a - b) (c - a)}} . x^{\frac{c + a}{(a - b) (b - c)}} . x^{\frac{a + b}{(b - c) (c - a)}} \\ = x^{\frac{(b + c) (b - c) + (c + a) (c - a) + (a + b) (a - b)}{(a - b) (b - c) (c - a)}} \\ = x^{\frac{(b^{2} - c^{2}) + (c^{2} - a^{2}) + (a^{2} - b^{2})}{(a - b) (b - c) (c - a)}} \\ = x^{0} \\ = 1 \end{aligned}

$\eqalign{ & {x^{\frac{{b + c}}{{\left( {a - b} \right)\left( {c - a} \right)}}}}.{x^{\frac{{c + a}}{{\left( {a - b} \right)\left( {b - c} \right)}}}}.{x^{\frac{{a + b}}{{\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {b + c} \right)\left( {b - c} \right) + \left( {c + a} \right)\left( {c - a} \right) + \left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^{\frac{{\left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right) + \left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}}} \cr & = {x^0} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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