What is factor 1+27e^{3x} ?

The solution to factor 1+27e^{3x} is (3e^x+1)(9e^{2x}-3e^x+1)

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factor 1+27e^{3x}

factor

1 + 27 e^{3 x}

(3 e^{x} + 1) (9 e^{2 x} - 3 e^{x} + 1)

Solution steps

1 + 27 e^{3 x}

Rewrite

27 e^{3 x}

(3 e^{1 x})^{3}

= 1 + (3 e^{1 \cdot x})^{3}

Rewrite

1

1^{3}

= (3 e^{1 x})^{3} + 1^{3}

Apply Sum of Cubes Formula:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(3 e^{1 x})^{3} + 1^{3} = (3 e^{1 x} + 1) ((3 e^{1 x})^{2} - 3 e^{1 x} + 1)

= (3 e^{1 x} + 1) ((3 e^{1 x})^{2} - 3 e^{1 x} + 1)

Simplify

(3 e^{1 \cdot x} + 1) ((3 e^{1 \cdot x})^{2} - 3 e^{1 \cdot x} + 1) : (3 e^{x} + 1) (9 e^{2 x} - 3 e^{x} + 1)

= (3 e^{x} + 1) (9 e^{2 x} - 3 e^{x} + 1)

(5 x + y)^{2} + (5 x + y) - 42

t^{2} - 1.4 t + 0.49

(m - 3)^{3} + (j - k)^{3}

x^{2} + 11 - 36

e^{2 x} \frac{d y}{d x} + e^{2 x} 2 y

What is factor 1+27e^{3x} ?
The solution to factor 1+27e^{3x} is (3e^x+1)(9e^{2x}-3e^x+1)