Question

prove that(x^a-b)^a+b (x^b-c) (x^c-a)^c + a = 1​

Answer 1

Solution :

We have to show that [ x^(a-b)]^(a+b) × [ x^(b-c)]^(b+c) × [ x^(c-a)]^(c+a) = 1

x^(a-b)]^(a+b) × [ x^(b-c)]^(b+c) × [ x^(c-a)]^(c+a)

> x^(a-b)(a+b) × x^(b-c)(b+c) × x^(c-a)(c+a)

> x^(a²-b²) × x^(b²-c²) × x^(c²-a²)

> x^(a²-b²+b²-c²+c²-a²)

> x^(a²+b²+c²-a²-b²-c²)

> x^⁰

> 1

Hence Proved

________________________________

Answer 2

Appropriate Question :

$\sf Prove \: that \: {x}^{(a - b)(a + b)} \cdot {x}^{(b - c)(b + c)} \cdot \: {x}^{(c - a)(c - a)} = 1$

Here, the dot represents the multiplication sign

Solution :

$\sf {x}^{(a - b)(a + b)} \cdot {x}^{(b - c)} \cdot \: {x}^{(c - a)(c - a)} = 1$

Now using the identity (a - b)(a + b) = a² - b² we get

$\sf {x}^{ ({a}^{2} - {b}^{2} )} \cdot {x}^{(b - c)} \cdot {x}^{( {c}^{2} - {a}^{2}) } = 1$

Now according the Product Law we know that

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

So by using this law we get

$\sf \leadsto {x}^{ {a}^{2} - {b}^{2} + b² - c² - {c}^{2} - {a}^{2} } = 1$

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

We know that anything to the power 0 is 1

So, x⁰ = 1

$\sf \leadsto{x}^{0} = 1$

$\sf \leadsto1 = 1$

LHS = RHS

$\green{ \underbrace{ \blue{HENCE, \: PROVED}}}$

Know More :

• a⁰ = 1

• a^m = aⁿ → m = n

• aⁿ = bⁿ → a = b

• (ab)ⁿ = aⁿbⁿ

• (a/b)ⁿ = aⁿ/bⁿ

Regards

# BeBrainly