\displaystyle \textbf{ALGEBRAIC IDENTITIES}

\displaystyle \text{We know from the distributive law that}

\displaystyle a(b+c)=ab+ac

\displaystyle \text{and}

\displaystyle (a+b)c=ac+bc.

\displaystyle \text{Using the distributive law, we can derive several useful algebraic}
\displaystyle \text{identities. These identities help us expand, simplify and factorise}
\displaystyle \text{algebraic expressions quickly.}
\displaystyle \\

\displaystyle \textbf{PRODUCT OF TWO BINOMIALS}

\displaystyle \text{1. }(x+a)(x+b)=x^2+(a+b)x+ab

\displaystyle \text{2. }(x+a)(x-b)=x^2+(a-b)x-ab

\displaystyle \text{3. }(x-a)(x+b)=x^2+(b-a)x-ab

\displaystyle \text{4. }(x-a)(x-b)=x^2-(a+b)x+ab

\displaystyle \text{In general,}

\displaystyle (x+a)(x+b)=x^2+(a+b)x+ab.
\displaystyle \\

\displaystyle \textbf{PRODUCT OF THE SUM AND DIFFERENCE OF TWO TERMS}

\displaystyle (a+b)(a-b)=a^2-b^2.

\displaystyle \text{Therefore,}

\displaystyle a^2-b^2=(a+b)(a-b).

\displaystyle \text{This identity is also known as the difference of two squares.}
\displaystyle \\

\displaystyle \textbf{SQUARES OF BINOMIALS}

\displaystyle \text{1. }(a+b)^2=a^2+2ab+b^2

\displaystyle \text{2. }(a-b)^2=a^2-2ab+b^2

\displaystyle \text{3. }a^2+b^2=(a+b)^2-2ab

\displaystyle \text{4. }a^2+b^2=(a-b)^2+2ab

\displaystyle \text{5. }(a+b)^2+(a-b)^2=2(a^2+b^2)

\displaystyle \text{6. }(a+b)^2-(a-b)^2=4ab

\displaystyle \text{7. }a^2+b^2= \frac{1}{2}\left\{(a+b)^2+(a-b)^2\right\}

\displaystyle \text{8. }ab= \frac{1}{4}\left\{(a+b)^2-(a-b)^2\right\}

\displaystyle \text{9. }(a+b)^2=(a-b)^2+4ab

\displaystyle \text{10. }(a-b)^2=(a+b)^2-4ab

\displaystyle \text{11. }\left(a+\frac{1}{a}\right)^2 =a^2+\frac{1}{a^2}+2,\qquad a\neq0

\displaystyle \text{12. }\left(a-\frac{1}{a}\right)^2 =a^2+\frac{1}{a^2}-2,\qquad a\neq0
\displaystyle \\

\displaystyle \textbf{PERFECT SQUARE TRINOMIALS}

\displaystyle \text{We know that}

\displaystyle a^2+2ab+b^2=(a+b)^2

\displaystyle \text{and}

\displaystyle a^2-2ab+b^2=(a-b)^2.

\displaystyle \text{Therefore, a trinomial of either of these forms is called a perfect}
\displaystyle \text{square trinomial because it can be expressed as the square of a binomial.}

\displaystyle \text{For example,}

\displaystyle x^2+6x+9=x^2+2(x)(3)+3^2=(x+3)^2.

\displaystyle \text{Similarly,}

\displaystyle x^2-10x+25=x^2-2(x)(5)+5^2=(x-5)^2.
\displaystyle \\

\displaystyle \textbf{SQUARES OF TRINOMIALS}

\displaystyle \text{1. }(a+b+c)^2 =a^2+b^2+c^2+2ab+2bc+2ca

\displaystyle =a^2+b^2+c^2+2(ab+bc+ca).

\displaystyle \text{2. }(a+b-c)^2 =a^2+b^2+c^2+2ab-2bc-2ca

\displaystyle =a^2+b^2+c^2+2(ab-bc-ca).

\displaystyle \text{3. }(a-b+c)^2 =a^2+b^2+c^2-2ab-2bc+2ca

\displaystyle =a^2+b^2+c^2+2(-ab-bc+ca).

\displaystyle \text{4. }(a-b-c)^2 =a^2+b^2+c^2-2ab+2bc-2ca

\displaystyle =a^2+b^2+c^2+2(-ab+bc-ca).

\displaystyle \text{5. }a^2+b^2+c^2 =(a+b+c)^2-2(ab+bc+ca)

\displaystyle \text{6. }2(ab+bc+ca) =(a+b+c)^2-(a^2+b^2+c^2)
\displaystyle \\

\displaystyle \textbf{CUBES OF BINOMIALS}

\displaystyle \text{1. }(a+b)^3 =a^3+3a^2b+3ab^2+b^3

\displaystyle =a^3+b^3+3ab(a+b).

\displaystyle \text{2. }(a-b)^3 =a^3-3a^2b+3ab^2-b^3

\displaystyle =a^3-b^3-3ab(a-b).

\displaystyle \text{3. }a^3+b^3 =(a+b)^3-3ab(a+b)

\displaystyle \text{4. }a^3-b^3 =(a-b)^3+3ab(a-b)
\displaystyle \\

\displaystyle \textbf{SUM AND DIFFERENCE OF TWO CUBES}

\displaystyle \text{1. }a^3+b^3=(a+b)(a^2-ab+b^2)

\displaystyle \text{2. }a^3-b^3=(a-b)(a^2+ab+b^2)
\displaystyle \\

\displaystyle \textbf{SPECIAL CUBIC IDENTITIES}

\displaystyle \text{For }a\neq0,

\displaystyle \text{1. }\left(a+\frac{1}{a}\right)^3 =a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)

\displaystyle \text{Therefore,}

\displaystyle a^3+\frac{1}{a^3} =\left(a+\frac{1}{a}\right)^3 -3\left(a+\frac{1}{a}\right).

\displaystyle \text{2. }\left(a-\frac{1}{a}\right)^3 =a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)

\displaystyle \text{Therefore,}

\displaystyle a^3-\frac{1}{a^3} =\left(a-\frac{1}{a}\right)^3 +3\left(a-\frac{1}{a}\right).
\displaystyle \\

\displaystyle \textbf{IMPORTANT IDENTITY FOR THREE VARIABLES}

\displaystyle a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca).

\displaystyle \text{Since}

\displaystyle a^2+b^2+c^2-ab-bc-ca =\frac{1}{2}\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\},

\displaystyle \text{we also have}

\displaystyle a^3+b^3+c^3-3abc

\displaystyle =\frac{1}{2}(a+b+c) \left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}.

\displaystyle \text{If }a+b+c=0,\text{ then}

\displaystyle a^3+b^3+c^3=3abc.
\displaystyle \\

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