Question

prove that please
I will Mark it as the best ​

Answer 1

Step-by-step explanation:

(2^a-b)^a+b*(2^b-c)^b+c*(2^c-a)^c+a

2^a^2-b^2+b^2-c^2+c^2-a^2=1

2^0=1

hence proved

Answer 2

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

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

$= 2^{a^2-b^2}*2^{b^2-c^2}*2^{c^2-a^2}$

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

$= 2^0$

$= 1 = RHS$