We know, that by Distributive law $\ a\left(b+c\right)=ab+ac\ and\ \left(a+b\right)c=ab+ac$. Using this we can derive formula for a lot of algebraic expressions. I have deliberately not given the proof details as it is just simple application of the above law. $\\$

To find the product of $(x\pm{}a)(x\pm{}b)$

1. $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$
2. $\left(x+a\right)\left(x-b\right)=x^2+\left(a-b\right)x-ab$
3. $\left(x-a\right)\left(x+b\right)=x^2+\left(b-a\right)x-ab$
4. $\left(x-a\right)\left(x-b\right)=x^2-\left(a+b\right)x+ab$

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Product of Sum and Difference of two terms

1. $\left(a+b\right)\left(a-b\right)=a^2-b^2$

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Squares of Binomials

1. ${(a+b)}^2=\ a^2+2ab+b^2$
2. ${(a-b)}^2=\ a^2-2ab+b^2$
3. $a^2+b^2=\ {(a+b)}^2-2ab$
4. $a^2+b^2=\ {(a+b)}^2-2ab$
5. $a^2+b^2=\ {(a-b)}^2+2ab$
6. $ab=\ \frac{1}{4}\{\left({a+b)}^2-\ {\left(a-b\right)}^2\right\}$
7. $a^2+b^2=\ \frac{1}{2}\{{\left(a+b\right)}^2-\ {\left(a-b\right)}^2\}$
8. ${(a+b)}^2=\ {(a-b)}^2+4ab$
9. ${\left(a+\frac{1}{a}\right)}^2=\ a^2+\frac{1}{a^2}+2$
10. ${\left(a-\frac{1}{a}\right)}^2=\ a^2+\frac{1}{a^2}-2$

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Perfect Square Trinomial

We know that $a^2+2ab+b^2=\ {(a+b)}^2 \ \ and \ \ a^2-2ab+b^2=\ {\left(a-b\right)}^2$, therefore every algebraic expression in the above form is called a perfect square.

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Squares of Trinomials

1. ${(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)$
2. ${(a+b-c)}^2=\left(a^2+b^2+c^2\right)+2(ab-bc-ca)$
3. ${(a-b+c)}^2=\left(a^2+b^2+c^2\right)+2(-ab-bc+ca)$
4. ${(a-b-c)}^2=\left(a^2+b^2+c^2\right)+2(-ab+bc-ca)$
5. $\left(a^2+b^2+c^2\right)={(a+b+c)}^2-2(ab+bc+ca)$
6. $2\left(ab+bc+ca\right)={\left(a+b+c\right)}^2-\left(a^2+b^2+c^2\right)$

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Cubes of Binomials

1. ${(a+b)}^3=\ a^3+b^3+3ab(a+b)$
2. ${(a-b)}^3=\ a^3-b^3-3ab(a-b)$
3. ${a^3+b^3=\ (a+b)}^3-3ab(a+b)$
4. ${a^3-b^3=\ (a-b)}^3+3ab(a-b)$
5. ${\left(a+\frac{1}{a}\right)}^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)$
6. ${\left(a+\frac{1}{a}\right)}^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)$
7. ${\left(a-\frac{1}{a}\right)}^3=a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)$