Show that the sum of the third roots of 1 is zero? Thank you!

1 Answer
George C.
Jan 15, 2018

See explanation...

Explanation:

Suppose #alpha, beta# and #gamma# are the zeros of a monic cubic polynomial. Then that polynomial can be written:

#(x-alpha)(x-beta)(x-gamma)=x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

Note that the third roots of #1# are the three zeros of the cubic polynomial:

#x^3+0x^2+0x-1#

So equating the coefficients of #x^2#, we find #alpha+beta+gamma=0#

Bonus

If you would like to explicitly find all the third roots of #1# you can do it like this:

#0 = x^3-1#

#color(white)(0) = (x-1)(x^2+x+1)#

#color(white)(0) = (x-1)((x+1/2)^2+3/4)#

#color(white)(0) = (x-1)((x+1/2)^2-(sqrt(3)/2i)^2)#

#color(white)(0) = (x-1)((x+1/2)-sqrt(3)/2i)((x+1/2)+sqrt(3)/2i)#

#color(white)(0) = (x-1)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)#

So the three roots are:

#1#, #" "-1/2+sqrt(3)/2i" "# and #" "-1/2-sqrt(3)/2i#

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