What is expand e^{x+y+z}(e^z-e^{x+y}) ?

The solution to expand e^{x+y+z}(e^z-e^{x+y}) is e^{2z+x+y}-e^{2x+2y+z}

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expand e^{x+y+z}(e^z-e^{x+y})

expand

e^{x + y + z} (e^{z} - e^{x + y})

e^{2 z + x + y} - e^{2 x + 2 y + z}

Solution steps

e^{x + y + z} (e^{z} - e^{x + y})

Apply the distributive law:

a (b - c) = ab - a c

a = e^{x + y + z}, b = e^{z}, c = e^{x + y}

= e^{x + y + z} e^{z} - e^{x + y + z} e^{x + y}

= e^{z} e^{x + y + z} - e^{x + y + z} e^{x + y}

Simplify

e^{z} e^{x + y + z} - e^{x + y + z} e^{x + y} : e^{2 z + x + y} - e^{2 x + 2 y + z}

= e^{2 z + x + y} - e^{2 x + 2 y + z}

ln (2) \cdot 2^{2 (- 4 x^{3} - 5 x^{2}) + 1} (- 12 x^{2} - 10 x)

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

(sec (x) - tan (x))^{2} d x

sin (42 0^{\circ})

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

What is expand e^{x+y+z}(e^z-e^{x+y}) ?
The solution to expand e^{x+y+z}(e^z-e^{x+y}) is e^{2z+x+y}-e^{2x+2y+z}