Answer by Mr.Wizard for How to simplify Sqrt[1/x] Sqrt[x]?

That is only true if C (in your short example) is positive, therefore you must instruct Mathematica to make such assumptions:

FullSimplify[expr, Λ> 0]

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You could also do this with $Assumptions:

$Assumptions = {Λ> 0};FullSimplify[expr]

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Blackbird suggests:

PowerExpand[expr] // FullSimplify

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Where:

expr = (9 + 9 E^((4 t Sqrt[Λ])/Sqrt[3]) - 12 y^2 Λ -      12 z^2 Λ+ 6 y^2 Sqrt[1/Λ] Λ^(3/2) +     6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 +     y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 +     2 x^2 Λ (-6 +        3 Sqrt[1/Λ] Sqrt[Λ] + (y^2 +           z^2) Λ) +     6 E^((2 t Sqrt[Λ])/Sqrt[3]) (3 - 2 z^2 Λ+        x^2 (-2 + Sqrt[1/Λ] Sqrt[Λ]) Λ+        y^2 (-2 + Sqrt[1/Λ] Sqrt[Λ]) Λ+        z^2 Sqrt[1/Λ] Λ^(3/2)))/(9 +     9 E^((4 t Sqrt[Λ])/Sqrt[3]) - 12 y^2 Λ -      12 z^2 Λ+ 6 y^2 Sqrt[1/Λ] Λ^(3/2) +     6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 +     y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 +     2 x^2 Λ (-6 +        3 Sqrt[1/Λ] Sqrt[Λ] + (y^2 +           z^2) Λ) -      6 E^((2 t Sqrt[Λ])/Sqrt[3]) (-3 +        x^2 Sqrt[1/Λ] Λ^(3/2) +        y^2 Sqrt[1/Λ] Λ^(3/2) +        z^2 Sqrt[1/Λ] Λ^(3/2)));