What is the answer to m=[((f(6)+f(2)))/((f(3)))]^{f(1)} ?

The answer to m=[((f(6)+f(2)))/((f(3)))]^{f(1)} is True

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m=[((f(6)+f(2)))/((f(3)))]^{f(1)}

m = [\frac{( f ( 6 ) + f ( 2 ) )}{( f ( 3 ) )}]^{f (1)}

True

Solution steps

m = [\frac{( f ( 6 ) + f ( 2 ) )}{( f ( 3 ) )}]^{f (1)}

Simplify

[\frac{( f ( 6 ) + f ( 2 ) )}{( f ( 3 ) )}]^{f (1)} : (\frac{f ( 6 ) + f ( 2 )}{f ( 3 )})^{f (1)}

m = (\frac{f ( 6 ) + f ( 2 )}{f ( 3 )})^{f (1)}

Both sides are equal

True

\frac{( x + a )}{a} - \frac{( x + b )}{b} = 1

x, 3 k = 2 x - 1

x, x (1 + p^{2}) = 1

\frac{5}{1} - \frac{1}{3} \frac{x}{1} = 15

x, y - 6 = 9 x - 30

What is the answer to m=[((f(6)+f(2)))/((f(3)))]^{f(1)} ?
The answer to m=[((f(6)+f(2)))/((f(3)))]^{f(1)} is True