What is expand (ace+bdf)^4 ?

The solution to expand (ace+bdf)^4 is e^4a^4c^4+4e^3a^3c^3bdf+6e^2a^2c^2b^2df^2+4eacb^3df^3+b^4df^4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (ace+bdf)^4

expand

(a ce + b df)^{4}

e^{4} a^{4} c^{4} + 4 e^{3} a^{3} c^{3} b df + 6 e^{2} a^{2} c^{2} b^{2} d f^{2} + 4 e a c b^{3} d f^{3} + b^{4} d f^{4}

Solution steps

(a ce + b df)^{4}

= (e a c + b df)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = a ce, b = b df

= i = 0 \sum 4 (i 4) (a ce)^{(4 - i)} (b df)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (a ce)^{4} (b df)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (a ce)^{3} (b df)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (a ce)^{2} (b df)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (a ce)^{1} (b df)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (a ce)^{0} (b df)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (a ce)^{4} (b df)^{0} : e^{4} a^{4} c^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (a ce)^{3} (b df)^{1} : 4 e^{3} a^{3} c^{3} b df

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (a ce)^{2} (b df)^{2} : 6 e^{2} a^{2} c^{2} b^{2} d f^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (a ce)^{1} (b df)^{3} : 4 e a c b^{3} d f^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (a ce)^{0} (b df)^{4} : b^{4} d f^{4}

= e^{4} a^{4} c^{4} + 4 e^{3} a^{3} c^{3} b df + 6 e^{2} a^{2} c^{2} b^{2} d f^{2} + 4 e a c b^{3} d f^{3} + b^{4} d f^{4}

9 x - 3 x + 5 (1 + 2 x = 7 x + 3 (x - 4))

2 ((5 y + 9) (2 y - 2))

\frac{1}{lo g _{10} ( x ) ( 8 + x )}

t^{2} H (t - 4) d t

x^{(a - b) (a^{2} + ab + b^{2})}

What is expand (ace+bdf)^4 ?
The solution to expand (ace+bdf)^4 is e^4a^4c^4+4e^3a^3c^3bdf+6e^2a^2c^2b^2df^2+4eacb^3df^3+b^4df^4