Question

Show that:
(x^a²+b²/x^ab)^a+b (x^b²+c²/x^bc) (x^c²+a²/x^ac) = x²(a³+b³+c³)

HOTS(Higher Order Thinking Skills).
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Answer 1

$\large\underline{\sf{Solution-}}$

Prove that:

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

Consider LHS,

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

We know that,

• $\sf \dfrac{p^m}{p^n}=p^{m-n}$

So,

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

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

We know that,

• $\sf (p^m)^n=p^{mn}$

So,

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

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

So,

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

On regrouping,

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

So,

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

We also know that,

• $\sf p^m\times p^n=p^{m+n}$

So,

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

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

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

So,

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

Taking 2 common,

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

LHS = RHS,

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

Hence, proved.