What is the proof that -
$α^2$ + $β^2$ = $(α+β)^2$ - 2αβ
$α^3$ + $β^3$ = $(α+β)^3$ - 3αβ(α+β)
$α^4+β^4$ = ($α^3+β^3$)(α+β) - αβ($α^2+β^2$)
(α+β)4 = α4 + 6α3β + $4α^2β^2$ + $6αβ^3$ + $β^4$
= $α^4$ + $β^4$ + $6αβ(α^2+β^2)$ + $4α^2β^2$
$α^5 + β^5$ = ($α^3+β^3$)($α^2+β^2$)-($α^2β^3+α^3β^2$) =($α^3+β^3$)($α^2+β^2$)-($α^2β^2$)(α+β)
mod(α-β) = ($(α+β)^2$ - 4αβ)
$α^2-β^2$ = (α+β)(α-β)
$α^3-β^3$ = $(α-β)^3$ + 3αβ(α-β) = (α-β)($(α+β)^2$-αβ)
$α^4-β^4$ = ($α^2+β^2$)($α^2-β^2$)
= (α+β)(α-β)($α^2+β^2$)
This website simply says that the terms on the left are somehow equal to the ones on the right, but it doesn't explain what it did to come to that conclusion (did it complete the squares? Is there some general axiom for this kind of thing?). Please, if you're familiar with this topic, can you give a step-by-step explanation? Also, if you know the use of symmetry (i.e. could it be used to plot graphs, etc), tell me. I need to know quadratic equations like the back of my hand. Thanks :)
Here's a sample question btw - If α and β are the roots of 3x$^2$ - 4x - 1 = 0, find:
i. α + β
ii. αβ
iii. $α^2$ + $β^2$
iv. 1/α + 1/β
v. α/β + β/α
vi. $α^3$ + $β^3$
vii. α - β
viii. 1/α+1 + 1/β+1
(P.S. Sorry for the long post)
Kinda weird to answer my own question but i think all i have to do is rewrite the equations in terms of $\alpha+\beta$ and $\alpha\beta$ to make them easier to work with. For $\alpha^3$+$\beta^3$, you rewrite it as the sum of two cubes ($\alpha+\beta$)($\alpha^2-\alpha\beta+\beta^2$). The expression in the second parentheses can be rewritten as ($\alpha^2+\beta^2-\alpha\beta$), and since $\alpha^2+\beta^2$ = ($\alpha+\beta)^2-2\alpha\beta$ (the square of a sum minus the $2\alpha\beta$), we should have something like ($(\alpha+\beta)^2$ - $2\alpha\beta - \alpha\beta$) in the second parentheses. Substitute $\alpha\beta$ with $\frac{c}{a}$ and $\alpha+\beta$ with $\frac{-b}{a}$ and voila, it should be easier to simplify now. You just have to rewrite it in terms of $\alpha\beta$ and $\alpha+\beta$. Still, i'm not familiar with how to solve expressions like $(a+b)^4$, $(a+b)^5$, $(a+b)^6$...$(a+b)^{62}$ without doing it the manual and tedious way.