Question

show that:- (x^a+b)^2 × (x^b+c)^2 × (x^c+a)^2 / (x^a × x^b × x^c) = 1​

Answer 1

Answer:

(x^a+b) ^ 2 x (x ^ b+c) ^ 2 x (x^c+a)^ 2

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x^a x^b x^c

= {(x)^ab} ^2 {(x)^bc}^ 2 { (x)^ ca} ^2

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{(3x)²}3

=. (x)⁶

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(x)⁶

= 1

Answer 2

To prove that:

$\dfrac{(x ^{a + b}) ^{2} \times ({x}^{b + c})^{2} \times ( {x}^{c + a})^{2} }{( {x}^{a} \times {x}^{b} \times {x}^{c}) ^{4} } = 1$

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we know that,

$( {a}^{m}) ^{n} = {a}^{mn}$

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$\dfrac{( {x}^{2(a + b)}) \times ( {x}^{2(b + c)}) \times {x}^{2(c + a)}) }{( {x}^{a} \times {x}^{b} \times {x}^{c}) ^{4} }$

$\dfrac{ ({x}^{2a + 2b}) \times ( {x}^{2b + 2c}) \times ({x}^{2c + 2a}) }{( {x}^{a} \times {x}^{b} \times {x}^{c}) ^{4} }$

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

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$\dfrac{ {x}^{2a + 2b + 2c+2b + 2c + 2a} }{( {x}^{a + b + c})^{4} }$

$\dfrac{ {x}^{4a + 4b + 4c} }{ {x}^{4(a + b + c)} }$

$\dfrac{ {x}^{4(a + b + c)} }{ {x}^{4(a + b + c)} } = 1$

Hence verified

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Important points to note:

$\dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

${a}^{n} \times {b}^{n} = {ab}^{n}$

${a}^{ - m} = \frac{1}{ {a}^{m} }$

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