What is expand (-1+sqrt(3))^4 ?

The solution to expand (-1+sqrt(3))^4 is 28-16sqrt(3)

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expand (-1+sqrt(3))^4

expand

(- 1 + 3)^{4}

28 - 163

Decimal Notation

0.28718 \dots

Solution steps

(- 1 + 3)^{4}

Apply binomial theorem:

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

a = - 1, b = 3

= i = 0 \sum 4 (i 4) (- 1)^{(4 - i)} (3)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (- 1)^{4} (3)^{0} : 1

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (- 1)^{3} (3)^{1} : - 43

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (- 1)^{2} (3)^{2} : 18

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (- 1)^{1} (3)^{3} : - 123

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (- 1)^{0} (3)^{4} : 9

= 1 - 43 + 18 - 123 + 9

1 - 43 + 18 - 123 + 9 = 28 - 163

= 28 - 163

(3 x^{2} + 6 x y^{2}) d x + (6 x^{2} y + 4 y^{3}) d y

\frac{1}{( t ^{2} - 1 ) ^{2}}

(- 3, 2, - 2) y (- 2, 3, - 4)

(x + \frac{1}{2}) (x + \frac{1}{4})

(cot (θ) + 1)^{2}

What is expand (-1+sqrt(3))^4 ?
The solution to expand (-1+sqrt(3))^4 is 28-16sqrt(3)