What is expand (a+b)^4-1 ?

The solution to expand (a+b)^4-1 is a^4+4a^3b+6a^2b^2+4ab^3+b^4-1

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expand (a+b)^4-1

expand

(a + b)^{4} - 1

a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} - 1

Solution steps

(a + b)^{4} - 1

Apply binomial theorem:

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

a = a, b = b

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

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} a^{4} b^{0} : a^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} a^{3} b^{1} : 4 a^{3} b

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} a^{2} b^{2} : 6 a^{2} b^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} a^{1} b^{3} : 4 a b^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} a^{0} b^{4} : b^{4}

= a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} - 1

sin (x^{2}) (x^{2} + 1)

- 2 z^{3} + 6 z^{2} + 4 z

\frac{( 3 t ^{3} + 6 t ^{2} - 15 t )}{( 3 t )}

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

a^{8} + 3 a^{4} + 4

What is expand (a+b)^4-1 ?
The solution to expand (a+b)^4-1 is a^4+4a^3b+6a^2b^2+4ab^3+b^4-1