Question

write the value of {xa/xb}a+b × {xb/xc}b+c × {xc/xa}c+a​

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

write the value of (X^a/x^b)^(a+b) * (x^b/x^c)^(b+c) * (x^c/x^a)^(c+a).

|| ★★ FORMULA USED ★★ ||

• (X^a/x^b) = x^(a-b)
• (x^a)^b = (x)^ab
• (a+b)(a-b) = (a² - b²)
• x^a * x^b * x^c = x^(a+b+c)
• x^0 = 1

|| ✰✰ ANSWER ✰✰ ||

☞ (X^a/x^b)^(a+b) * (x^b/x^c)^(b+c) * (x^c/x^a)^(c+a)

using (X^a/x^b) = x^(a-b) First we get,

☞ {x^(a-b)}^(a+b)} * {x^(b-c)}^(b+c)} * {x^(c-a)}^(c+a)}

Using (x^a)^b = (x)^ab now, we get,

☞ (x)^{(a-b)(a+b)} * (x)^{(b-c)(b+c)} * (x)^{(c-a)(c+a)}

Using (a+b)(a-b) = (a² - b²) Now,

☞ (x)^{a² - b²} * (x)^{b²-c²} * (x)^{c²-a²}

using x^a * x^b * x^c = x^(a+b+c) now,

☞ (x)^[ a² - b² + b² - c² + c² - a²]

☞ (x)^0

using x^0 = 1 in last ,

☛ 1 (Ans.)

Answer 2

Question :-

Write the value of :

$\sf \large \red{\sf \bigg\{ \frac{ {x}^{a} }{ {x}^{b} } \bigg\} ^{a + b} }\times \green{ \bigg \{\frac{ {x}^{b} }{ {x}^{c} } \bigg\} ^{ b + c }}\times \pink{ \bigg\{ \frac{ {x}^{c} }{ {x}^{a} } \bigg \} ^{c + a }}$

Answer :-

→ 0

Solution :-

We have :

$\sf \large\green{\sf \bigg\{ \frac{ {x}^{a} }{ {x}^{b} } \bigg\} ^{a + b} }\times \orange{ \bigg \{\frac{ {x}^{b} }{ {x}^{c} } \bigg\} ^{ b + c }}\times \purple{ \bigg\{ \frac{ {x}^{c} }{ {x}^{a} } \bigg \} ^{c + a }}\\ \: \\ \because \boxed{\sf \frac{ {x}^{y} }{ {x}^{z} } = {x}^{(y - z)} } \\ \therefore \\ \\ \to \sf \bold\{ \blue{ { {x}^{(a - b)} }^{(a + b)} } \} \times \{ { \red{{x}^{(b - c)} }^{(b + c)}\}} \times \{{ \pink{{x}^{(c - a)} }^{(c + a)} }\} \\ \\ \sf \: \: \: \because \boxed{ \sf { {x}^{a} }^{b} = {x}^{ab} } \\ \\ \sf\to \bold \: \sf \{ \pink{ { x }^{(a + b)(a - b)} } \} \times \{ \green{ {x}^{(b - c)(b + c)} } \}\times \{ \orange{ {x}^{(c + a)(c - a)} } \}$

$\sf \because \boxed{ (x + y)( x - y) = {x}^{2} - {y}^{2} } \\ \\ \therefore \\ \to \sf \{ { x }^{ \red{{a}^{2} - {b}^{2} } } \} \times \{ \blue{ {x}^{( {b}^{2} - {c}^{2}) } } \}\times \{ \orange{ {x}^{( {c}^{2} - {a}^{2} )} } \}$

We know that, power of same base be added .

$\to \sf \large {x}^{ \red{ {a}^{2} } - {b}^{2} + \green{ {b}^{2} - } {c}^{2} + \pink{{c}^{2} - {a}^{2} }} \\ \\ \to \large \boxed{\sf {x}^{0} = 1}$