Question

solve please I will mark you as brilliants​

Answer 1

ANSWER:

To Prove:

$\:\:\bullet\:\:\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}=x^{2(a^3+b^3+c^3)}$

Solution:

We need to prove that,

$\implies\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}=x^{2\left(a^3+b^3+c^3\right)}$

Taking LHS,

$\implies\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}$

We know that,

$\hookrightarrow \dfrac{p^m}{p^n}=p^{m-n}$

So,

$\implies\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}$

$\implies\left(x^{a^2+b^2-ab}\right)^{a+b}\left(x^{b^2+c^2-bc}\right)^{b+c}\left(x^{c^2+a^2-ca}\right)^{c+a}$

We know that,

$\hookrightarrow (p^m)^n=p^{mn}$

So,

$\implies\left(x^{a^2+b^2-ab}\right)^{a+b}\left(x^{b^2+c^2-bc}\right)^{b+c}\left(x^{c^2+a^2-ca}\right)^{c+a}$

$\implies\left(x^{a^2(a)+a^2(b)+b^2(a)+b^2(b)-ab(a)-ab(b)}\right)\left(x^{b^2(b)+b^2(c)+c^2(b)+c^2(c)-bc(b)-bc(c)}\right)\left(x^{c^2(c)+c^2(a)+a^2(c)+a^2(a)-ca(c)-ca(a)}\right)$

So,

$\implies\left(x^{a^3+a^2b+ab^2+b^3-a^2b-ab^2}\right)\left(x^{b^3+b^2c+bc^2+c^3-b^2c-bc^2}\right)\left(x^{c^3+ac^2+a^2c+a^3-ac^2-a^2c}\right)$

On regrouping,

$\implies\left(x^{a^3+b^3+(a^2b-a^2b)+(ab^2-ab^2)}\right)\left(x^{b^3+c^3+(b^2c-b^2c)+(bc^2-bc^2)}\right)\left(x^{c^3+a^3+(ac^2-ac^2)+(a^2c-a^2c)}\right)$

So,

$\implies\left(x^{a^3+b^3}\right)\left(x^{b^3+c^3}\right)\left(x^{c^3+a^3}\right)$

We also know that,

$\hookrightarrow p^m\times p^n=p^{m+n}$

So,

$\implies\left(x^{a^3+b^3}\right)\left(x^{b^3+c^3}\right)\left(x^{c^3+a^3}\right)$

$\implies\left(x^{a^3+b^3+b^3+c^3+c^3+a^3}\right)$

$\implies x^{\left(a^3+a^3+b^3+b^3+c^3+c^3\right)}$

So,

$\implies x^{\left(2a^3+2b^3+2c^3\right)}$

Taking 2 common,

$\implies \bf x^{2\left(a^3+b^3+c^3\right)}=RHS$

As, LHS = RHS,

HENCE PROVED, that,

$\bf\large\implies\left(\dfrac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\dfrac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\dfrac{x^{c^2+a^2}}{x^{ca}}\right)^{c+a}=x^{2\left(a^3+b^3+c^3\right)}$