What is ((1-sqrt(3))^4)/(16)+sqrt(3)-1 ?

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

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simplify ((1-sqrt(3))^4)/(16)+sqrt(3)-1

simplify

\frac{( 1 - 3 ) ^{4}}{16} + 3 - 1

\frac{3}{4}

Decimal Notation

0.75

Solution steps

\frac{( 1 - 3 ) ^{4}}{16} + 3 - 1

Apply the fraction rule:

a = \frac{ab}{b}

= \frac{( 1 - 3 ) ^{4}}{16} + \frac{3 \cdot 16}{16} - 1

Convert

- 1

to fraction

: \frac{- 16}{16}

= \frac{( 1 - 3 ) ^{4}}{16} + \frac{3 \cdot 16}{16} + \frac{- 16}{16}

\frac{( 1 - 3 ) ^{4}}{16} + \frac{3 \cdot 16}{16} + \frac{- 16}{16} = \frac{( 1 - 3 ) ^{4} + 3 \cdot 16 - 16}{16}

= \frac{( 1 - 3 ) ^{4} + 3 \cdot 16 - 16}{16}

Simplify

(1 - 3)^{4} + 3 \cdot 16 - 16 : 12

= \frac{12}{16}

Cancel

\frac{12}{16} : \frac{3}{4}

= \frac{3}{4}

(y^{\frac{3}{4}} - 2 x^{\frac{3}{4}})^{3}

\frac{3}{13} \cdot (7 - 3) + \frac{4}{13} (0.25 \cdot 3) + \frac{14}{13} (- 3)

\frac{1 - 3 i}{1 + 3 i}

\frac{1}{x + 4}

t \cdot (1 - H (t - 8))

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