What is ((3-isqrt(3))/3)((2+isqrt(3))/3) ?

The solution to ((3-isqrt(3))/3)((2+isqrt(3))/3) is 1+1/(3sqrt(3))i

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simplify ((3-isqrt(3))/3)((2+isqrt(3))/3)

simplify

(\frac{3 - i 3}{3}) (\frac{2 + i 3}{3})

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

Solution steps

(\frac{3 - i 3}{3}) (\frac{2 + i 3}{3})

Simplify

(\frac{3 - i 3}{3}) (\frac{2 + i 3}{3}) : \frac{3 - i 3}{3} \cdot \frac{2 + i 3}{3}

= \frac{3 - i 3}{3} \cdot \frac{2 + i 3}{3}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( 3 - i 3 ) ( 2 + i 3 )}{3 \cdot 3}

Apply FOIL method

(3 - i 3) (2 + i 3) : i 3 + 9

= \frac{i 3 + 9}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{i 3 + 9}{9}

Rewrite

i 3 + 9

in standard complex form:

9 + 3 i

= \frac{9 + 3 i}{9}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{9}{9} + \frac{3}{9} i

Simplify

\frac{9}{9} + \frac{3}{9} i : 1 + \frac{1}{3 \cdot 3} i

= 1 + \frac{1}{3 \cdot 3} i

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

\frac{1}{i ^{- 14}}

lo g_{6} (\frac{1}{36}) = x

2 y^{3} + 1 y^{2} - 2 y - 1

x^{4} - 3 x^{3} + 4 x

x \cdot 2 x^{2}

What is ((3-isqrt(3))/3)((2+isqrt(3))/3) ?
The solution to ((3-isqrt(3))/3)((2+isqrt(3))/3) is 1+1/(3sqrt(3))i