The value of xa-b2xb-c2xc-a2-xa-xb-xc4 is

The correct option is C 1 The given expression is: #160; (x ^ (a+b) )^ 2 (x ^ (c+b) )^ 2 (x ^ (a+c) )^ 2 ( x ^ a . x ^ b . x ^ c )^ 4 = (x ^ ...

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Byju's Answer

Standard XII

Mathematics

Graphical Method of Solving Linear Programming Problems

The value of ...

Question

The value of $\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}}$ is

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Solution

The correct option is C $1$
The given expression is:
$\frac{{(x}^{(a + b)})^{2} {(x}^{(c + b)})^{2} {(x}^{(a + c)})^{2}}{(x^{a} . x^{b} . x^{c})^{4}}$
$= \frac{{(x}^{(2 a + 2 b)}) {(x}^{(2 c + 2 b)}) {(x}^{(2 a + 2 c)})}{(x^{a + b + c})^{4}}$
$= \frac{{(x}^{(2 a + 2 b)}) {(x}^{(2 c + 2 b)}) {(x}^{(2 a + 2 c)})}{(x^{4 a + 4 b + 4 c})}$
$= \frac{{(x}^{(2 a + 2 b + 2 b + 2 c + 2 c + 2 a)})}{(x^{4 a + 4 b + 4 c})}$
$= \frac{{(x}^{(4 a + 4 b + 4 c)})}{(x^{4 a + 4 b + 4 c})} = 1$

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Similar questions

Q. Show that

\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} x^{b} x^{c})}^{4}} = 1

Prove that :

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

(ii) ${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b} = 1$

Q. Find the value of the expression:

\frac{(X^{a + b})^{2} \times (X^{b + c})^{2} \times (X^{c + a})^{2}}{(X^{a} X^{b} X^{c})^{4}}

, if

x = 2, a = 1, b = 2, c = 3

Q. Simplify :

{(\frac{x^{a}}{x^{b}})}^{a^{2} + a b + b^{2}} \times {(\frac{x^{b}}{x^{c}})}^{b^{2} + b c + c^{2}} \times {(\frac{x^{c}}{x^{a}})}^{c^{2} + c a + a^{2}}

State true or false

Simplified form of

(\frac{x^{a}}{x^{b}})^{a^{2} + a b + b^{2}} \times (\frac{x^{b}}{x^{c}})^{b^{2} + b c + c^{2}} \times (\frac{x^{c}}{x^{a}})^{c^{2} + c a + a^{2}}

is 1

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The value of (xa+b)2(xb+c)2(xc+a)2(xa.xb.xc)4 is

The correct option is C 1The given expression is: (x(a+b))2(x(c+b))2(x(a+c))2(xa.xb.xc)4=(x(2a+2b))(x(2c+2b))(x(2a+2c))(xa+b+c)4=(x(2a+2b))(x(2c+2b))(x(2a+2c))(x4a+4b+4c)=(x(2a+2b+2b+2c+2c+2a))(x4a+4b+4c)=(x(4a+4b+4c))(x4a+4b+4c)=1

The value of $\frac{{(x^{a + b})}^{2} {(x^{b + c})}^{2} {(x^{c + a})}^{2}}{{(x^{a} . x^{b} . x^{c})}^{4}}$ is