xa-xb c- xb-xc a- xc-xa b-

The correct option is B 1 ( x^ax^b )^c× ( x^bx^c )^a× ( x^cx^a )^b =x^ac bc× x^ab ac× x^bc ab =x^ac bc+ab ac+bc ab =x^0=1 ...

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Byju's Answer

Standard IX

Mathematics

Properties of Measures of Central Tendency

xa/xb c× xb/x...

Question

${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b}$ =_______.

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Solution

The correct option is B $1$
${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b}$

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

$= x^{a c - b c + a b - a c + b c - a b}$

$= x^{0} = 1$

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Similar questions

Q. Prove that,

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

Q. Simplify:

{(\frac{x^{b}}{x^{c}})}^{a}

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

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

Q. Simplify

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

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

is equal to

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

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(xaxb)c×(xbxc)a×(xcxa)b=_______.

The correct option is B 1(xaxb)c×(xbxc)a×(xcxa)b=xac−bc×xab−ac×xbc−ab=xac−bc+ab−ac+bc−ab=x0=1

${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b}$ =_______.

The correct option is B $1$
${(\frac{x^{a}}{x^{b}})}^{c} \times {(\frac{x^{b}}{x^{c}})}^{a} \times {(\frac{x^{c}}{x^{a}})}^{b}$

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

$= x^{a c - b c + a b - a c + b c - a b}$

$= x^{0} = 1$