Question

(x^a-b)^a+b(x^b-c)b+c(x^c-a)c+a=1​

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

• Here we go through L.H.S

⟹(x^a-b)^a+b×(x^b-c)^b+c×(x^c-a)^c+a

⟹(x)^a²-b²×(x)^b²-c²+(x)^c²-a²

⟹(x)^a²-b²+b²-c²+c²-a²

⟹(x)^0=>X=1

• using above identity
• a²-b²=(a+b)(a-b)

$\bold\orange{\boxed{Answer\:x=1}}$

Answer 2

$\huge\tt{SOLUTION:}$

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Solving the Left Hand Side-

➩(x^a-b)^a+b×(x^b-c)^b+c×(x^c-a)^c+a=1

➩(x)^a²-b²×(x)^b²-c²+(x)^c²-a²=1

➩(x)^a²-b²+b²-c²+c²-a²= 1

➩(x)^0=1

➩x=1 = 1

$\tt{L.H.S=R.HS}$

Hence,

Proved

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