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Solve the cubic equation $z^3+6z^2+12z+16=0$

and show the three solutions on an Argand diagram

HINT: $(a+b)^3$

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    @Henning Makholm Indeed... Deleting my previous comments.2011-12-30

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Okay, lets proceed with your hint then.

$z^3+6z^2+12z+16=0$ $(z+2)^3+8=0$ $(z+4)(z^2+2z+4)=0$ $(z+4)[(z+1)^2+3]=0$

So you obtain $z=-4, -1+i\sqrt{3},-1-i\sqrt{3}$

It should be easy to represent these on an argand diagram.

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    why not just take cube roots of minus eight and shift by minus 2?2011-12-30