Question

if you want to be one of the brainliest so please answer this question I lied you fastly in the list of branliest

Answer 1

Answer:

Step-by-step explanation:

Given :

To prove :

$(\frac{x^{a^{2}}}{x^{b^{2}}})^{\frac{1}{a+b}}+(\frac{x^{b^{2}}}{x^{c^{2}}})^{\frac{1}{b+c}}+(\frac{x^{c^{2}}}{x^{a^{2}}})^{\frac{1}{a+c}}=1$

Solution :

We know that,

$x^0=1$ (anything to the power 0 is 1)

$\frac{x^a}{x^b}=x^{a-b}$

Also,

$a^2 - b^2 = (a+b)(a-b)$

So,

LHS = $(\frac{x^{a^{2}}}{x^{b^{2}}})^{\frac{1}{a+b}}+(\frac{x^{b^{2}}}{x^{c^{2}}})^{\frac{1}{b+c}}+(\frac{x^{c^{2}}}{x^{a^{2}}})^{\frac{1}{a+c}}$

⇒ $({x^{a^{2} - b^{2}}})^{\frac{1}{a+b}}+({x^{b^{2}-c^{2}}})^{\frac{1}{b+c}}+({x^{c^{2}-a^{2}}})^{\frac{1}{a+c}}$

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

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

⇒ $x^{(a-b)} \times x^{(b-c)} \times x^{(c-a)} = x^{a-b+b-c+c-a} = x^0 = 1$ = RHS

Hence proved,.