Question

${({2}^{x} - 4)}^{3} + {({4}^{x} - 2)}^{3} = {({4}^{x} + {2}^{x} - 6 )}^{3}$
​Sum of all real values of x is ?

Answer 1

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

Given that

$\rm :\longmapsto\:{({2}^{x} - 4)}^{3} + {({4}^{x} - 2)}^{3} = {({4}^{x} + {2}^{x} - 6 )}^{3}$

can be rewritten as

$\rm :\longmapsto\:{({2}^{x} - 4)}^{3} + {({4}^{x} - 2)}^{3} = {({4}^{x} + {2}^{x} - 4 - 2)}^{3}$

$\rm :\longmapsto\:{({2}^{x} - 4)}^{3} + {({4}^{x} - 2)}^{3} = { \bigg(({4}^{x} - 2) + ({2}^{x} - 4) \bigg)}^{3}$

Let assume that

$\rm :\longmapsto\: {2}^{x} - 4 = a$

and

$\rm :\longmapsto\: {4}^{x} - 2 = b$

So, given equation can be rewritten as

$\rm :\longmapsto\: {a}^{3} + {b}^{3} = {(a + b)}^{3}$

$\rm :\longmapsto\: \cancel {a}^{3} + \cancel {b}^{3} = \cancel {a}^{3} +\cancel {b}^{3} + 3ab(a + b)$

$\rm :\longmapsto\:3ab(a + b) = 0$

$\rm :\implies\:a = 0 \: \: or \: \: b = 0 \: \: or \: a + b = 0$

Consider,

$\rm :\longmapsto\:a = 0$

$\rm :\longmapsto\: {2}^{x} - 4 = 0$

$\rm :\longmapsto\: {2}^{x} = 4$

$\rm :\longmapsto\: {2}^{x} = {2}^{2}$

$\bf :\implies\:x = 2 - - - (1)$

Consider,

$\rm :\longmapsto\:b = 0$

$\rm :\longmapsto\: {4}^{x} - 2 = 0$

$\rm :\longmapsto\: {2}^{2x} = 2$

$\rm :\longmapsto\: {2}^{2x} = {2}^{1}$

$\rm :\longmapsto\:2x = 1$

$\bf :\implies\:x = \dfrac{1}{2} - - - (2)$

Consider,

$\rm :\longmapsto\:a + b = 0$

$\rm :\longmapsto\: {2}^{x} - 4 + {4}^{x} - 2 = 0$

$\rm :\longmapsto\: {2}^{x}+ {4}^{x} = 2 + 4$

$\rm :\longmapsto\: {2}^{x}+ {4}^{x} = {2}^{1} + {4}^{1}$

On comparing, we get

$\bf\implies \:x = 1 - - - (3)$

So, from equation (1), (2) and (3), we concluded that

$\bf :\longmapsto\:x = 2 \: \: or \: \: \dfrac{1}{2} \: \: or \: \: 1$

Hence,

Sum of all real values of x is

$\rm \: = \: \: 1 + 2 + \dfrac{1}{2}$

$\rm \: = \: \: 3+ 0.5$

$\rm \: = \: \: 3.5$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{\boxed{ \bf{ \: Option \: (D) \: is \: correct}}}$

Identities Used :-

$\rm :\longmapsto\: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

$\rm :\longmapsto\: {a}^{x} = {a}^{y} \implies \: x = y$