Question

show that [x^1/a-b]^1^a-c.[x^1/b-c]^1/b-a.[x^1/c-a]^1/c-b=1 where a is not equal to b is not equal to c .
Please help me......​

Answer 1

Answer:

(x)^1/(a-b)(a-c)*(x)^1/(b-c)(b-a)*(x)^1/(c-a)(c-b)

(x)^1/(a-b)(a-c)+1/(b-c)(b-a)+1/(c-a)(c-b)

(x)^1/(a-b)(a-c)+1/(b-c)+(1/(b-a)-1/(c-a)

x^1/(a-b)(a-c)+1/(b-c)(c-a+a-b/(b-a)(c-a)

x^1/(a-b)(a-c)+1/(c-b)/(b-c)(b-a)(c-a)

x^/(a-b)(a-c)+1/(b-a)(c-a)

x^1/(a-b)(c-a+a-c/(c-a)

0/x^a-b=1

unless x=0

Step-by-step explanation:

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Answer 2

Answer:

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