# Factorise z^3+1 into a linear and quadratic factor. Let y=(1+i√3)/2. Show that y is a cube root of -1. Show that y^2=y-1. Find the value of (1-y)^6.

To factorise z^{3}+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z^{3}+1=0.

We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)^{3}+1=0 since (-1)^{3}=-1.

Therefore, (z+1) is a factor. The other factor is quadratic. We know, by inspection, that the quadratic factor will look like: (z^{2}+kx+1) where k is a constant. This is because z^{3}+1 can be written as a product of its factors: z^{3}+1=(z+1)(z^{2}+kx+1). The coefficient of the z^{3} term and constant term are 1 so we know the coefficient of the z^{2 }and constant term in the quadratic. By inspection we can also see that k=-1.

The second part of the questions asks us to show that y^{3}=-1. There are several approaches for this. You can square y to obtain y^{2} then multiply it by y again to obtain y^{3} and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y^{3} then convert back into Cartesian form. The final method is as follows:

We want to show that y^{3}=-1. Earlier we wrote: z^{3}=-1 which leads to z^{3}+1=(z+1)(z^{2}-z+1)=0. We know that if y is a cubic root of -1 then it cannot be equal to -1, so for the above to be true then (z^{2}-z+1) must be equal to 0. We simply solve the quadratic (z^{2}-z+1)=0 using the quadratic formula. The formula gives us that z=(1+i√3)/2 after taking the positive result. We note that this is equal to y. Since y=x then y is a cube root of -1.

The third part of the question asks us to show that y^{2}=y-1. Using our working from earlier, we know that y is a root of z^{2}-z+1=0 so y^{2}-y+1=0. Rearranging this we have y^{2}=y-1 as required. Alternatively you can find y^{2} then show that it is equal to y-1.

The final part of the question is to find the value of (1-y)^{6}. Instinctively most people would use the binomial expansion but there is a far more elegant solution. From the previous part we have y^{2}=y-1. If we multiply both sides by -1 then:

(-y^{2} )=1-y

Raising both sides to the power 6 as the question required and using power laws:

(-y^{2})^{6}=(1-y)^{6}

Why would we do this though? Maybe you can see why; there is a piece of information from earlier that we can use, if not, keep following and you will see that this method is very clever.

(-y)^{6*2}=(y)^{12}=(1-y)^{6}

The negative sign disappears as squaring a number makes it positive.

We can manipulate this further:

(y)^{12}=(y)^{3*4}=(y^{3})^{4}

Earlier we had (y^{3})=-1 and so ultimately the value of (1-y)^{6}=(-1)^{4}=1.