Rumus-Rumus Aljabar

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a² + b² = (a + b)² − 2ab

a³ + b³ = (a + b)³ − 3ab(a + b)

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3(a² + b²)(a + b) = 2(a³ + b³) + (a + b)³

$x^2+{\displaystyle \frac{1}{x^2}}={(x+{\displaystyle \frac{1}{x}})}^2-2$

BUKTI

$$\begin{array}{rl}{(x+{\displaystyle \frac{1}{x}})}^2& =x^2+2x\left({\displaystyle \frac{1}{x}}\right)+{\displaystyle \frac{1}{x^2}}\\ & =x^2+2+{\displaystyle \frac{1}{x^2}}\\ x^2+{\displaystyle \frac{1}{x^2}}& ={(x+{\displaystyle \frac{1}{x}})}^2-2\end{array}$$

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$x^3+{\displaystyle \frac{1}{x^3}}={(x+{\displaystyle \frac{1}{x}})}^3-3(x+{\displaystyle \frac{1}{x}})$

BUKTI

$$\begin{array}{rl}{(x+{\displaystyle \frac{1}{x}})}^3& =x^3+3x^2\left({\displaystyle \frac{1}{x}}\right)+3x{\left({\displaystyle \frac{1}{x}}\right)}^2+{\left({\displaystyle \frac{1}{x}}\right)}^3\\ & =x^3+3x+3x\left({\displaystyle \frac{1}{x^2}}\right)+{\displaystyle \frac{1}{x^3}}\\ & =x^3+3x+3\left({\displaystyle \frac{1}{x}}\right)+{\displaystyle \frac{1}{x^3}}\\ & =x^3+{\displaystyle \frac{1}{x^3}}+3(x+{\displaystyle \frac{1}{x}})\\ x^3+{\displaystyle \frac{1}{x^3}}& ={(x+{\displaystyle \frac{1}{x}})}^3-3(x+{\displaystyle \frac{1}{x}})\end{array}$$

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