Question

(x^a/x^b)^a+b*(x^b/x^c)^b+c*(x^c/x^a)^c+a​

Answer 1

Answer:

1

check the answer in the picture !!!

Answer 2

$\bigg(\dfrac{x^{a} }{x^{b} } \bigg)^{a+b}\times \bigg(\dfrac{x^{b} }{x^{c} } \bigg)^{b+c}\times \bigg(\dfrac{x^{c} }{x^{a} } \bigg)^{c+a}$

Notice that here, the base values are the same.

According to the exponent law :-

• $\dfrac{a^{m} }{a^{n} } =a^{m-n}$ ('a' being the common base)

Hence, here the fractions can be expressed in this way.

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

Here, since the terms are whole powered by another power,

• $(a^{m})^{n} = a^{m\times n}$

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

Let us take the powers alone.

For term one :-

The powers are not multiplied yet, the product is not yet obtained form

=> (a-b)(a+b)

For term 2 :-

Here too, the powers are yet to multiplied,

=> (b-c)(b+c)

For term 3 :-

The same case for term 3's powers

=> (c-a)(c+a)

According to an identity :-

The difference of 2 square numbers, say 'y' and 'z' are expressed as :-

• y² - z² = (y-z)(y+z)

Here too the power's are in this form. Hence without actually performing multiplication, we can say that the powers will be :-

Term 1 :-

a² - b²

Term 2 :-

b² - c²

Term 3 :-

c² - a²

Now that we have solved the powers, by putting them in the their respective places, we get,

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

According to another exponent law,

• $a^{m} \times a^{n} = a^{m+n}$

Hence by applying it,

$(x^{(a^{2}-b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2})})$

By removing the brackets we get,

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

All the powers cancel out.

We are left with :-

$x^{0}$

We know that,

• $a^{0} = 1$

The answer thus will be :-

$x^{0} = 1$

Therefore the final answer is 1