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 z³+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z³+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: (z²+kx+1) where k is a constant. This is because z³+1 can be written as a product of its factors: z³+1=(z+1)(z²+kx+1). The coefficient of the z³ term and constant term are 1 so we know the coefficient of the z² 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 y³=-1. There are several approaches for this. You can square y to obtain y² then multiply it by y again to obtain y³ and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y³ then convert back into Cartesian form. The final method is as follows:We want to show that y³=-1. Earlier we wrote: z³=-1 which leads to z³+1=(z+1)(z²-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 (z²-z+1) must be equal to 0. We simply solve the quadratic (z²-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 y²=y-1. Using our working from earlier, we know that y is a root of z²-z+1=0 so y²-y+1=0. Rearranging this we have y²=y-1 as required. Alternatively you can find y² 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 y²=y-1. If we multiply both sides by -1 then:(-y² )=1-yRaising both sides to the power 6 as the question required and using power laws:(-y²)⁶=(1-y)⁶Why 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)⁶²=(y)¹²=(1-y)⁶The negative sign disappears as squaring a number makes it positive.We can manipulate this further:(y)¹²=(y)³4=(y³)⁴Earlier we had (y³)=-1 and so ultimately the value of (1-y)⁶=(-1)⁴=1.