Developing a formula from cube equation

[elimenohpee]

Homework Statement

Given that the equation $$x^{3} + ax^{2} + bx + c = 0$$ has the roots $$r_{1}, r_{2}, r_{3}$$, develop formulae for $$r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3}$$

Homework Equations

The Attempt at a Solution

Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
$$(x-r_{1})(x-r_{2})(x-r_{3})=0$$

 

[LCKurtz]

Homework Statement

Given that the equation $$x^{3} + ax^{2} + bx + c = 0$$ has the roots $$r_{1}, r_{2}, r_{3}$$, develop formulae for $$r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3}$$

Homework Equations

The Attempt at a Solution

Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
$$(x-r_{1})(x-r_{2})(x-r_{3})=0$$

Multiply those out and see what happens.

 

[elimenohpee]

Is that really all I need to do? I mean when I multiply it out, I get:

$$x^{3}-cx^{2}-bx^{2}-ax^{2}+bcx+acx+abx-abc=0$$

The question says to develop formulae, all three formulas are within this equation...is there anything else I can really say here? lol

 

[LCKurtz]

Homework Statement

Given that the equation $$x^{3} + ax^{2} + bx + c = 0$$ has the roots $$r_{1}, r_{2}, r_{3}$$, develop formulae for $$r_{1} + r_{2} + r_{3}, r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}, r_{1}r_{2}r_{3}$$

Homework Equations

The Attempt at a Solution

Not really sure where to attempt to start this problem. Only thing I know is that the roots of the equation imply
$$(x-r_{1})(x-r_{2})(x-r_{3})=0$$

Multiply those out and see what happens.

Is that really all I need to do? I mean when I multiply it out, I get:

$$x^{3}-cx^{2}-bx^{2}-ax^{2}+bcx+acx+abx-abc=0$$

The question says to develop formulae, all three formulas are within this equation...is there anything else I can really say here? lol

? When you multiply out

$$(x-r_{1})(x-r_{2})(x-r_{3})=0$$

you won't have any a, b, or c. The question is how does what you get relate to the original cubic you started with:

$$x^{3} + ax^{2} + bx + c = 0$$

 

[elimenohpee]

wow, sorry I see now. Forgive what I just wrote, it is late and I wasn't thinking correctly.

When I multiply out, I get:

$$x^{3} -x^{2}(r_{1}+r_{2}+r_{3}) + x(r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}) -r_{1}r_{2}r_{3}=0$$

then you can equate coefficients with the original cubic expression to get:

$$x^{3}$$ term:$$1-1=0$$
$$x^{2}$$ term: $$-(r_{1}+r_{2}+r_{3})=a$$
$$x$$ term:$$r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}=b$$
initial term:$$r_{1}r_{2}r_{3} = -c$$