Verified answer

Since α is a root of x^2 = 2x-1 ( in other words x^2 - 2x + 1 = 0),

if you substitute α instead of x in the LHside of the eq, RHside should be zero.

ie, α^2 -2α + 1 = 0 ------(a)

If you multiply (a) by α, you get,

α^3 -2α^2 + α = 0 , or α^3 = 2α^2 - α (b)

Now since

α^2 -2α + 1 = 0, 2 x (α^2 -2α + 1) = 0

So if you sustract 2 x (α^2 -2α + 1) [which is actually zero] from the RHside of (b), its LHside should remain the same, ie,

α^3 = 2α^2 - α - 2 x (α^2 -2α + 1)

Simplifying,

α^3 = 2α^2 - α - 2α^2 + 4α - 2 = 3α - 2

PS. It was easier for the earlier 3 people since the equation was easily solvable. But that would have been difficult otherwise. For instance if the question was;

Given that α is the root of the equation x^2 = 2x+1, show that α^3 = 5α +2 ?

Then the eq. is not factorisable, and the roots will not be whole numbers but involving decimals ! But you will not have problems with my method!!

Try and see for practice !

Root=solution

Turn the equation given into a quadratic.

x^2 - 2x + 1 = 0

Solve the quadratic. You should get x=1. This is the root of the equation (typically quadratics will have more than one answer, but in this case there is only one). A is the root of the equation, so that means that a=x and therefore a=1.

Now substitute 1 for a in the equation a^3 = 3a - 2

1^3 = 3(1) - 2

Simplify each side.

Left side: 1^3 = 1

Right side: 3(1) - 2 = 1

Now the overall equation is simply

1=1

If you show your work (the steps you took to solve the equation), it should be enough to get you full credit.

if x^2 = 2x -1 then

x^2 -2x +1 =0 or x =1

so α = 1

α^3 = 3α -2 is true

This is also true

α^999 = 4α -3

x² = 2x - 1

x² - 2x + 1 = 0

(x - 1)² = 0

x - 1 = 0

x = 1

α = x, so

α = 1

and

If α³ = 3a - 2,

then

(1)³ = 3(1) - 2

1 = 3 - 2

1 = 1

Q.E.D.