Question

please tell the answer fast​

Answer 1

Step-by-step explanation:

take denominator in numerrator

solve (a+b)(a-b)=a^2-b^2 and similarly for other 2 brackets also

now finally add all powers it will come zero

so x^0=1

HENCE PROVED

THANKS :)

Answer 2

Step-by-step explanation:

$= {( \frac{ {x}^{a} }{ {x}^{ - b} })}^{a - b} \times {( \frac{ {x}^{b} }{ {x}^{ - c} })}^{b - c} \times {( \frac{ {x}^{c} }{ {x}^{ - a} })}^{c - a} \\by \: using \: identity \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} \\ = ( { {x}^{a + b} )}^{a - b} \times ( { {x}^{b+ c} )}^{b - c} \times ( { {x}^{c + a } )}^{c - a}$

By using formula,

$({ {a}^{m} })^{n} = {a}^{m \times n}$

$(a - b)(a + b) = {a}^{2} - {b}^{2}$

$= {x}^{ ({a}^{2} - {b}^{2} ) } \times {x}^{ ({b}^{2} - {c}^{2} ) } \times {x}^{ ({c}^{2} - {a}^{2} ) }$

by using Identity,

${ a }^{m} \times {a}^{n} = {a}^{m +n}$

$= {x}^{ ({a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} )} \\ = {x}^{0} \\ = 1$

${a}^{0} = 1$

$Hence \: Shown$