Question

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Answer 1

Answer:

This is the answer.

I think it will help you .

Answer 2

Answer:

1=1

Step-by-step explanation:

Proof:

LHS-

We know, when the bases are same and we devide them; powers get subtracted.

$( { {3}^{a - b} )}^{a + b} \times ( { {3}^{b - c} )}^{b + c} \times ( { {3}^{c - a}) }^{c + a}$

Using a²-b²= (a+b)(a-b)

${3}^{ {a}^{2} - {b}^{2} } \times {3}^{ {b}^{2} - {c}^{2} } \times {3}^{ {c}^{2} - {a}^{2} }$

Now,

the bases are same, we are multiplying the them; powers get added

${3}^{ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} } \\ eliminate \: all \: opposites \\ {3}^{0}$

Any number raised to power 0=1

${3}^{0} = 1$

1=1; LHS=RHS

Hence proved