In a triangle ABC,if a

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(C) Given that,$\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$ $(i)$Using the sine rule for a triangle $ABC$,we have $\frac{a}{\sin A} = \frac{b}{\

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$2$

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(C) Given that,$\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$ $(i)$Using the sine rule for a triangle $ABC$,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,where $R$ is the circumradius.Thus,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into equation $(i)$:$\frac{(2R \sin A)^3 + (2R \sin B)^3 + (2R \sin C)^3}{\sin^3 A + \sin^3 B + \sin^3 C} = 8$$\frac{(2R)^3 (\sin^3 A + \sin^3 B + \sin^3 C)}{\sin^3 A + \sin^3 B + \sin^3 C} = 8$$(2R)^3 = 8 \implies 2R = 2 \implies R = 1$.Since $c = 2R \sin C = 2 \sin C$,and the maximum value of $\sin C$ is $1$ (as $C < 180^\circ$),the maximum value of $c$ is $2 	imes 1 = 2$.

In a triangle $ABC$,if $a < b < c$ and $\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$,then the maximum value of $c$ is

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