Identity 1: α² + β²

$$ \boxed{ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \phantom{.} } $$

Deriving the formula:

\begin{align*} \text{Since } (a + b)^2 & = a^2 + 2ab + b^2, \\ \\ (\alpha + \beta)^2 & = \alpha^2 + 2\alpha\beta + \beta^2 \\ (\alpha + \beta)^2 - 2\alpha \beta & = \alpha^2 + \beta^2 \\ \\ \therefore \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \end{align*}

Identity 2: (α - β)²

$$ \boxed{ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \phantom{.} } $$

Deriving the formula:

\begin{align*} \text{Since } (a - b)^2 & = a^2 - 2ab + b^2, \\ \\ (\alpha - \beta)^2 & = \alpha^2 - 2\alpha\beta + \beta^2 \\ & = \underbrace{\alpha^2 + \beta^2}_\text{Identity 1} - 2\alpha \beta \\ & = [(\alpha + \beta)^2 - 2 \alpha\beta] - 2\alpha\beta \\ & = (\alpha + \beta)^2 - 4 \alpha\beta \\ \\ \therefore (\alpha - \beta)^2 & = (\alpha + \beta)^2 - 4\alpha \beta \end{align*}

Identity 3: α³ + β³

Two versions:

$$ \boxed{ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \phantom{.} } $$

$$ \boxed{ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] \phantom{.} } $$

Converting from the first version to the second version:

\begin{align*} \alpha^3 + \beta^3 & = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \\ & = (\alpha + \beta) ( \underbrace{ \alpha^2 + \beta^2 }_\text{Identity 1} - \alpha \beta ) \\ & = (\alpha + \beta) [ (\alpha + \beta)^2 - 2 \alpha \beta - \alpha \beta] \\ & = (\alpha + \beta) [ (\alpha + \beta)^2 - 3 \alpha \beta] \end{align*}

Identity 4: α³ - β³

Two versions:

$$ \boxed{ \alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) \phantom{.} } $$

$$ \boxed{ \alpha^3 + \beta^3 = (\alpha - \beta)[(\alpha + \beta)^2 - \alpha\beta] \phantom{.} } $$

Converting from the first version to the second version:

\begin{align*} \alpha^3 - \beta^3 & = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) \\ & = (\alpha - \beta)( \underbrace{\alpha^2 + \beta^2}_\text{Identity 1} + \alpha \beta) \\ & = (\alpha - \beta) [(\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta ] \\ & = (\alpha - \beta) [(\alpha + \beta)^2 - \alpha\beta ] \end{align*}

Identity 5: α⁴ + β⁴

$$ \boxed{ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \phantom{.} } $$

Deriving the formula using identity 1:

\begin{align*} \text{Identity 1: } \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \\ \\ \text{Replace } \alpha \text{ by } \alpha^2 & \text{ and } \beta \text{ by } \beta^2, \\ \\ (\alpha^2)^2 + (\beta^2)^2 & = (\alpha^2 + \beta^2)^2 - 2(\alpha^2)(\beta^2) \\ \alpha^4 + \beta^4 & = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \end{align*}

Identity 6: α⁴ - β⁴

$$ \boxed{ \alpha^4 - \beta^4 = (\alpha^2 + \beta^2)(\alpha + \beta)(\alpha - \beta) \phantom{.} } $$

Deriving the formula using the identity $a^2 - b^2 = (a + b)(a - b)$:

\begin{align*} \alpha^4 - \beta^4 & = (\alpha^2)^2 - (\beta^2)^2 \\ & = (\alpha^2 + \beta^2) (\alpha^2 - \beta^2) \\ & = (\alpha^2 + \beta^2) (\alpha + \beta) (\alpha - \beta) \end{align*}