Math, asked by p8a6yantimaneshpl, 1 year ago

if a=b^2x,b=c^2y,c=a^2z,prove that xyz=1÷8

Answers

Answered by Zooni
82
a = b^2x
a = (c^2y)^2x
a = {(a^2z)^2y}^2x
a = a^8xyz
comparing powers on both sides…
1 = 8xyz
=>xyz = 1÷8
thus, LHS = RHS
hence proved.




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Answered by Trinankasarkar
3

Step-by-step explanation?

xyz=1/8

hence LHS=RHS

PROVED

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