What is expand (1-2sqrt(2))^4 ?

The solution to expand (1-2sqrt(2))^4 is 113-72sqrt(2)

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expand (1-2sqrt(2))^4

expand

(1 - 22)^{4}

113 - 722

Decimal Notation

11.17662 \dots

Solution steps

(1 - 22)^{4}

Apply binomial theorem:

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

a = 1, b = - 22

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (- 22)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (- 22)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (- 22)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (- 22)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (- 22)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (- 22)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (- 22)^{0} : 1

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (- 22)^{1} : - 82

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (- 22)^{2} : 48

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (- 22)^{3} : - 642

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (- 22)^{4} : 64

= 1 - 82 + 48 - 642 + 64

1 - 82 + 48 - 642 + 64 = 113 - 722

= 113 - 722

\frac{d x}{( x ^{2} + 1 ) ^{2}}

\frac{2 x + h}{x ^{2} ( x ^{2} + 2 x h + h ^{2} )}

3 ((x + 3) (x - 2) (x - 5))

(2 a b + 2 a c + a + b + 2 b c + c)^{2}

\frac{s + 4}{( s - 2 ) ( s + 1 )}

What is expand (1-2sqrt(2))^4 ?
The solution to expand (1-2sqrt(2))^4 is 113-72sqrt(2)