Simplify the following and leave your answer in standard form: (1-i)6 x (sqrt(3)cis(-30^))42i-1

First, convert $1-i$ to polar form $r(\cos\theta + i\sin\theta)$.
The modulus is $r = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The argument is $\theta = \arctan\left(\frac{-1}{1}\right) = -45^\circ$ (since $1-i$ is in the fourth quadrant).
So, $1-i = \sqrt{2}\text{cis}(-45^\circ)$.

Now, use De Moivre's Theorem to calculate $(1-i)^6$:

Since $-270^\circ$ is equivalent to $90^\circ$ (i.e., $-270^\circ + 360^\circ = 90^\circ$):

Convert back to rectangular form:

Use De Moivre's Theorem:

Convert to rectangular form:

Numerator $= (8i) \times \left(-\frac{9}{2} - i\frac{9\sqrt{3}}{2}\right)$

Since $i^2 = -1$:

The denominator is $2i-1$, which can be written as $-1+2i$.

Multiply the numerator and denominator by the conjugate of the denominator, which is $-1-2i$:

Calculate the denominator:

Calculate the numerator:

Group the real and imaginary parts:

Now, combine the numerator and denominator:

The final answer in standard form is $\boxed*{\frac{-36\sqrt{3} - 72}{5} + \frac{36 - 72\sqrt{3}}{5}i}$.