value of abc when x^a=y ; y^b =z ; z^c=x
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x^a=y, x^ab=y^b(multiplying the powers with b) x^ab=z(given that v^b= z) x^abc=z^c(multiplying the powers with c) x^abc=x(given that z^c=x) x^abc=x^1, abc=1 hence proved
Hope this helps u and good luck with the answer
Hope this helps u and good luck with the answer
sam12a13:
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Consider xa = y
⇒ (zc)a = y [Since zc=x]
⇒ zca = y
⇒ (yb)ca = y [Since yb = z]
⇒ ybca = y
⇒ yabc = y1
Comparing both the sides, we get
abc = 1
⇒ (zc)a = y [Since zc=x]
⇒ zca = y
⇒ (yb)ca = y [Since yb = z]
⇒ ybca = y
⇒ yabc = y1
Comparing both the sides, we get
abc = 1
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