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

The solution to expand (x^2+x^3)^4 is x^8+4x^9+6x^{10}+4x^{11}+x^{12}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

x^{8} + 4 x^{9} + 6 x^{10} + 4 x^{11} + x^{12}

Solution steps

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

Apply binomial theorem:

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

a = x^{2}, b = x^{3}

= i = 0 \sum 4 (i 4) (x^{2})^{(4 - i)} (x^{3})^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (x^{2})^{4} (x^{3})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (x^{2})^{3} (x^{3})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (x^{2})^{2} (x^{3})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (x^{2})^{1} (x^{3})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (x^{2})^{0} (x^{3})^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (x^{2})^{4} (x^{3})^{0} : x^{8}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (x^{2})^{3} (x^{3})^{1} : 4 x^{9}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (x^{2})^{2} (x^{3})^{2} : 6 x^{10}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (x^{2})^{1} (x^{3})^{3} : 4 x^{11}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (x^{2})^{0} (x^{3})^{4} : x^{12}

= x^{8} + 4 x^{9} + 6 x^{10} + 4 x^{11} + x^{12}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{7}{2}}} d t

y = 2 x^{2} - 1, (0, 3)

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 e^{- 2 t}

\int \frac{( x - 3 )}{( x + 4 ) ( x - 5 )} d x

\frac{d}{d x} (x^{3} - 5 x^{2} + 7 x - 3)

What is expand (x^2+x^3)^4 ?
The solution to expand (x^2+x^3)^4 is x^8+4x^9+6x^{10}+4x^{11}+x^{12}