What is expand (e^{2x}-5e^x+6=1)e^x ?

The solution to expand (e^{2x}-5e^x+6=1)e^x is e^{3x}-5e^{2x}+6e^x=1

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expand (e^{2x}-5e^x+6=1)e^x

expand

(e^{2 x} - 5 e^{x} + 6 = 1) e^{x}

e^{3 x} - 5 e^{2 x} + 6 e^{x} = 1

Solution steps

(e^{2 x} - 5 e^{x} + 6 = 1) e^{x}

= e^{x} (e^{2 x} - 5 e^{x} + 6 \cdot 1 =)

Distribute parentheses

e^{x} e^{2 x} + e^{x} (- 5 e^{x}) + e^{x} \cdot 6 = 1

Apply minus-plus rules

+ (- a) = - a

e^{x} e^{2 x} - 5 e^{x} e^{x} + 6 e^{x} = 1

Expand

e^{x} e^{2 x} - 5 e^{x} e^{x} + 6 e^{x} : e^{3 x} - 5 e^{2 x} + 6 e^{x}

e^{3 x} - 5 e^{2 x} + 6 e^{x} = 1

\frac{( x ^{2} + 1 ) ( x ^{2} - x + 1 )}{x ( x ^{2} + 2 )}

\frac{s ^{3} ( 5 t ^{2} - 6 t + 3 H ( t ) - 3 e ^{- 2 t} )}{3 ( s + 2 )}

2 (π - 2)

7 (tan^{2} (θ) - 1 + cot^{4} (θ))

\frac{s ^{3} + 2 s ^{2} + 5 s + 8}{( s ^{2} + 4 ) ( s ^{2} + 9 )}

What is expand (e^{2x}-5e^x+6=1)e^x ?
The solution to expand (e^{2x}-5e^x+6=1)e^x is e^{3x}-5e^{2x}+6e^x=1