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-y

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


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.


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

We can manipulate this further:


Earlier we had (y3)=-1 and so ultimately the value of (1-y)6=(-1)4=1.

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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.

Make y the subject of (y/x)+(2y/(x+4))=3

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