What is expand ((3^x+4^x)^2)/(12^x) ?

The solution to expand ((3^x+4^x)^2)/(12^x) is (3/4)^x+2+(4/3)^x

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expand ((3^x+4^x)^2)/(12^x)

expand

\frac{( 3 ^{x} + 4 ^{x} ) ^{2}}{1 2 ^{x}}

(\frac{3}{4})^{x} + 2 + (\frac{4}{3})^{x}

Solution steps

\frac{( 3 ^{x} + 4 ^{x} ) ^{2}}{1 2 ^{x}}

(3^{x} + 4^{x})^{2} = 9^{x} + 3^{x} \cdot 2^{1 + 2 x} + 1 6^{x}

= \frac{9 ^{x} + 3 ^{x} \cdot 2 ^{2 x + 1} + 1 6 ^{x}}{1 2 ^{x}}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{9 ^{x} + 3 ^{x} \cdot 2 ^{1 + 2 x} + 1 6 ^{x}}{1 2 ^{x}} = \frac{9 ^{x}}{1 2 ^{x}} + \frac{3 ^{x} \cdot 2 ^{1 + 2 x}}{1 2 ^{x}} + \frac{1 6 ^{x}}{1 2 ^{x}}

= \frac{9 ^{x}}{1 2 ^{x}} + \frac{3 ^{x} \cdot 2 ^{2 x + 1}}{1 2 ^{x}} + \frac{1 6 ^{x}}{1 2 ^{x}}

\frac{9 ^{x}}{1 2 ^{x}} = (\frac{3}{4})^{x}

\frac{3 ^{x} \cdot 2 ^{1 + 2 x}}{1 2 ^{x}} = 2

\frac{1 6 ^{x}}{1 2 ^{x}} = (\frac{4}{3})^{x}

= (\frac{3}{4})^{x} + 2 + (\frac{4}{3})^{x}

(2 x + y) (cos (t))

(e^{i θ} - 1) (e^{- i θ} + 1)

\frac{3 x + 5}{2 ( x + 2 )}

3 (2 cos (x) - x sin (x))

x \cdot (x + 1)

What is expand ((3^x+4^x)^2)/(12^x) ?
The solution to expand ((3^x+4^x)^2)/(12^x) is (3/4)^x+2+(4/3)^x