cubic equations (1 Viewer)

maths lover · May 18, 2011

i was trying to solve this question when i just couldn't find how.
without solving the polynomial X^3-X^2-X+1=0 WITH ROOTS A,B AND C. find the value of A^3+B^3+C^3

SpiralFlex · May 18, 2011

Factor theorem

Finding the factors of $x^3 - x^2 - x + 1$ ,

The possible factors are $1$ and $-1$ .

Let's try $1$ .

Substituting $1$ , $(1)^3 - (1)^2 - (1) + 1 = 0$

$\therefore$ $(x-1)$ is a factor.

Using division, you will get a quotient of $x^2 - 1$ which is $(x-1)(x+1)$ .

So $x^3-x^2-x+1$ in factored form is $(x-1)^2(x+1)$

$(x-1)^2(x+1) = 0$

So the roots $A, B, C$ are $1, 1, -1$

$A^3+B^3+C^3 = (1)^3 + (1)^3 + (-1)^3 = 1$

Aindan · May 18, 2011

Sub A, B, C into the equation (this works because they are the roots of the equation). Add all three equations together. Simplify using sum and product of roots.

AAEldar · May 18, 2011

Aindan said:
Sub A, B, C into the equation (this works because they are the roots of the equation). Add all three equations together. Simplify using sum and product of roots.

$x^3-x^2-x+1=0 \\ \text{Let the roots be A, B and C} \\ A+B+C=\frac{-b}{a} \\ =1 \\ AB+AC+BC=\frac{c}{a} \\ =-1 \\\\ \text{Since A, B and C are roots of the equation, sub for x} \\ A^3=A^2+A-1 \\ B^3=B^2+B-1 \\ C^3=C^2+C-1 \\ \text{Add the 3 equations} \\ A^3+B^3+C^3=(A^2+B^2+C^2)+(A+B+C)-3 \\ =(A+B+C)^2-2(AB+AC+BC)+(A+B+C) \\ =(1)^2-2(-1)+(1)-3 \\ =1 \\ \therefore A^3+B^3+C^3=1$

Theres the working for what Aindan said.

maths lover · May 18, 2011

Factor theorem

Finding the factors of ,

The possible factors are and .

Let's try .

Substituting ,

is a factor.

Using division, you will get a quotient of which is .

So in factored form is

So the roots are

isn't this solving but?

SpiralFlex · May 18, 2011

maths lover said:
isn't this solving but?

Not really, use Aiden's and AAEldar's method then. Which topic is this?

maths lover · May 18, 2011

i have it written as quadratic functions but i think it comes under polynomials.

SpiralFlex · May 18, 2011

maths lover said:
i have it written as quadratic functions but i think it comes under polynomials.

Which book? Also, what method does the worked examples show?

maths lover · May 18, 2011

that specific question is one that my teacher made, but it can be found in chapter 8 of the Cambridge textbook. excercise h

AAEldar · May 18, 2011

I'm pretty sure SpiralFlex's method is not the way the question asked. It gets the answer still, but the question said without solving i.e. without finding the roots.

You can do it the way I showed you, or you could expand $(A+B+C)^3$ , isolate $A^3$ , $B^3$ and $C^3$ and subtract the remaining terms, however that can be tedious.

SpiralFlex · May 18, 2011

AAEldar said:
I'm pretty sure SpiralFlex's method is not the way the question asked. It gets the answer still, but there question said without solving i.e. without finding the roots.

You can do it the way I showed you, or you could expand $(A+B+C)^3$ , isolate $A^3$ , $B^3$ and $C^3$ and subtract the remaining terms, however that can be tedious.

I will attempt to store your method in my memory and boot it up when it comes time for the HSC.

AAEldar · May 18, 2011

SpiralFlex said:
I will attempt to store your method in my memory and boot it up when it comes time for the HSC.

I learnt it the other day in 4U Polynomials, which was kinda surprising because it was so easy and simple and clean!

SpiralFlex · May 18, 2011

AAEldar said:
I learnt it the other day in 4U Polynomials, which was kinda surprising because it was so easy and simple and clean!

Seems crunchy. Partial fractions in 4U Poly for the win?

maths lover · May 18, 2011

You can do it the way I showed you, or you could expand , isolate , and and subtract the remaining terms

that is the way my teacher showed me but she showed us the pattern which we can use. but i posted the question on bos since i didn't want to remember the pattern.

Drongoski · May 18, 2011

Ainden/AAEldar's is the better way to do it.

gurmies · Jun 29, 2011

SpiralFlex said:
Factor theorem

Finding the factors of $x^3 - x^2 - x + 1$ ,

The possible factors are $1$ and $-1$ .

Let's try $1$ .

Substituting $1$ , $(1)^3 - (1)^2 - (1) + 1 = 0$

$\therefore$ $(x-1)$ is a factor.

Using division, you will get a quotient of $x^2 - 1$ which is $(x-1)(x+1)$ .

So $x^3-x^2-x+1$ in factored form is $(x-1)^2(x+1)$

$(x-1)^2(x+1) = 0$

So the roots $A, B, C$ are $1, 1, -1$

$A^3+B^3+C^3 = (1)^3 + (1)^3 + (-1)^3 = 1$

$\text{Too many unnecessary steps.} \,\, \text{Note that:} \\\\ x^{3}-x^{2}-x+1=x^{2}(x-1)-(x-1)\Leftrightarrow (x^{2}-1)(x-1) \,\, \text{then it follows that}\sum_{cyc}\alpha ^{3}=\boxed{1}.$

SpiralFlex · Jun 29, 2011

gurmies said:
$\text{Too many unnecessary steps.} \,\, \text{Note that:} \\\\ x^{3}-x^{2}-x+1=x^{2}(x-1)-(x-1)\Leftrightarrow (x^{2}-1)(x-1) \,\, \text{then it follows that}\sum_{cyc}\alpha ^{3}=\boxed{1}.$

I did that method so it would suit maths lover as he is in Preliminary.

cubic equations (1 Viewer)

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