Question

√x^a-b × √x^b-c × √x^c-a​

Answer 1

Answer:

1

Step-by-step explanation:

√x^(a-b+b-c+c-a)

=√x^(0)

=1

Answer 2

To solve:

√x^(a - b) . √x^(b - c) . √x^( c - a)

As,the base number is equal,we write:

→√x^[(a-b)+(b-c)+(c-a)]

→√x^{a-a+b-b+c-c}

→√x^{0}

→1

You might be confused how √x^0 can be 1.

There is a postulate which states that anything to the power of zero is one.

Proof:

To prove:

$x {}^{0} = 1$

Consider x^0,

$x {}^{0} \\ \\ = x {}^{1 - 1} \\ \\ = x {}^{1} .x {}^{ - 1} \\ \\ = x. \frac{1}{x} \\ \\ = 1$

Hence,proved