Question

Prove this question !!​

Answer 1

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

The above question is related to exponents and powers and we have to prove a certain equation which is as follows:

$\boxed{ \sf{ \bigg( \frac{ {x}^{a} }{ {x}^{b} } \bigg) {}^{a + b} \times \bigg( \frac{ {x}^{b} }{ {x}^{c} } \bigg) {}^{b + c} \times \bigg( \frac{ {x}^{c} }{ {x}^{a} } \bigg) {}^{c + a} = 1}}$

So, let's use different properties of the exponents as provided in the attachment and solve the question....

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LHS:

Firstly, let's use the Quotient property for inside of the parentheses and simplify it:

$\sf{ ({x}^{a - b} ) {}^{a + b} \times ( {x}^{b - c} ){}^{b + c} \times ({x}^{c - a} ) {}^{c + a} }$

Now multiplying the exponents of the terms:

$\sf{ {x}^{ {a}^{2} - {b}^{2} } \times {x}^{ {b}^{2} - {c}^{2} } \times {x}^{ {c}^{2} - {a}^{2} } }$

As we can see that the base of these terms are equal i.e. x, then we add the exponents with the same base:

$\sf{ { x}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } }$

Here a² get cancels with -a² and similarly b² and c² will get cancelled. Eventually the power is 0.

$\sf{ {x}^{0} }$

Anything to the power 0 is equals to 1. Thus, LHS is equals to RHS. Hence proved!

Answer 2

$\huge\underline{\underline{\bold{\red{A}\green{n}\orange{s}\pink{w}\blue{e}{r}}}}$

to prove your question equation,

let's start

• to solve this problem we use different properties of exponents

1. let's use quotient property

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

2. now , multiplying

= x^a^2 -b^2 × x^b^2 - c^2 × x^c^2-a^2

3. now , add

= x^a^2 -b^2 + b^2- c^2 + c^2- a^2

• here, the these exponents cancelled each another so, the power of these exponents is 0.
• anything in algebraic expression to the power 0 is equal to 1. it will be proved ✓
• so , the addition property is used in this question to prove this equation.

$\sf\large\underline\purple{answer by @itzfadedstar :)}$

Thanks