Given that x^3a = y^2b = z^4c = xyz,then 3ab + 4bc + 6ca = ?

Dear Vignesh

given

X^3a = Y^2b = Z^4c = XYZ

writing Y and Z in terms of X using above relation

Y=X^3a/2b, and Z=X^3a/4c

now XYZ = X*X^3a/2b*X^3a/4c

            = X^{1+(3a/2b)+3a/4c}

       XYZ    = X^{(4bc+6ac+3ab)/4bc}.........(1)

from given relation XYZ= X^3a........(2)

from (1) and (2)

X^3a=X^{(4bc+6ac+3ab)/4bc}

3a=(4bc+6ac+3ab)/4bc

hence 3ab + 4bc + 6ca = 12abc  Answer

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

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