Question

can someone explain how to use Pascal's triangle for cubing -1+((3i)^1/2)?

Answer 1

pascals triangle has rows of numbers that represent the coefficient of the terms in a bimonial expansion

Answer 2

the row for a ^3 is 1 3 3 1
then just define the ab parts to multiply to those

Answer 3

1 3 3 1 pascals row
-1 1 -1 1 (-1)^3,2,1,0
.... whatever the 3i 1/2s turn out to be from 0,1,2,3
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multiply down and add across

Answer 4

So would it be
\[1(1^{3}+\sqrt{3i})+3(1^{2}\times \sqrt{3i}^{0})+3(1^{1} \times \sqrt{3i}^{2})+1(1^{0} \times \sqrt{3i}^{2}\]
?

Answer 5

\[1(-1)^{3}(\sqrt{3i})^0~+3(-1)^{2}(\sqrt{3i})^{1}~+3(-1)^{1}(\sqrt{3i})^{2}+1(-1)^{0} (\sqrt{3i})^{3}\]
\[-1+3\sqrt{3i}~-3(3i)+3i\sqrt{3i}\]
yeah, something like that

Answer 6

ok thanks

Answer 7

youre welcome