\displaystyle \textbf{Chapter 4: Expansions}

\displaystyle \textbf{Concept Notes}

\displaystyle \textbf{1. Expansion}
\displaystyle \text{Expansion is the process of removing brackets and simplifying algebraic expressions.}

\displaystyle \textbf{2. Important Identities}
\displaystyle (a+b)^2=a^2+2ab+b^2
\displaystyle (a-b)^2=a^2-2ab+b^2
\displaystyle (a+b)(a-b)=a^2-b^2

\displaystyle \textbf{3. Expansion of }(a+b)^3
\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3

\displaystyle \textbf{4. Expansion of }(a-b)^3
\displaystyle (a-b)^3=a^3-3a^2b+3ab^2-b^3

\displaystyle \textbf{5. Useful Results from Cube Identities}
\displaystyle a^3+b^3=(a+b)^3-3ab(a+b)
\displaystyle a^3-b^3=(a-b)^3+3ab(a-b)

\displaystyle \textbf{6. Special Cases}
\displaystyle \left(a+\frac1a\right)^3=a^3+\frac1{a^3}+3\left(a+\frac1a\right),\ a\ne0
\displaystyle \left(a-\frac1a\right)^3=a^3-\frac1{a^3}-3\left(a-\frac1a\right),\ a\ne0

\displaystyle \textbf{7. If }a+b+c=0
\displaystyle a^3+b^3+c^3=3abc

\displaystyle \textbf{8. Expansion of }(x+a)(x+b)
\displaystyle (x+a)(x+b)=x^2+(a+b)x+ab

\displaystyle \textbf{9. Expansion of }(x+a)(x-b)
\displaystyle (x+a)(x-b)=x^2+(a-b)x-ab

\displaystyle \textbf{10. Expansion of }(x-a)(x+b)
\displaystyle (x-a)(x+b)=x^2+(b-a)x-ab

\displaystyle \textbf{11. Expansion of }(x-a)(x-b)
\displaystyle (x-a)(x-b)=x^2-(a+b)x+ab

\displaystyle \textbf{12. Expansion of }(a+b+c)^2
\displaystyle (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

\displaystyle \textbf{13. Expansion of }(a+b-c)^2
\displaystyle (a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca

\displaystyle \textbf{14. Expansion of }(a-b+c)^2
\displaystyle (a-b+c)^2=a^2+b^2+c^2-2ab-2bc+2ca

\displaystyle \textbf{15. Expansion of }(a-b-c)^2
\displaystyle (a-b-c)^2=a^2+b^2+c^2-2ab+2bc-2ca

\displaystyle \textbf{16. Important Results using }(a+b+c)^2
\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
\displaystyle \therefore ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}2

\displaystyle \textbf{17. Important Results using }(a+b-c)^2
\displaystyle (a+b-c)^2=a^2+b^2+c^2+2(ab-bc-ca)
\displaystyle \therefore ab-bc-ca=\frac{(a+b-c)^2-(a^2+b^2+c^2)}2

\displaystyle \textbf{18. Using Expansions}
\displaystyle \bullet\ \text{Evaluate algebraic expressions quickly using identities.}
\displaystyle \bullet\ \text{Find unknown quantities from given relations.}
\displaystyle \bullet\ \text{Find coefficients and constant terms in polynomial expansions.}
\displaystyle \bullet\ \text{Simplify complicated algebraic expressions.}
\displaystyle \bullet\ \text{Prove algebraic identities.}

\displaystyle \textbf{19. Special Products}
\displaystyle (x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc
\displaystyle (a+b)(a^2-ab+b^2)=a^3+b^3
\displaystyle (a-b)(a^2+ab+b^2)=a^3-b^3
\displaystyle (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc

\displaystyle \textbf{20. Examination Tips}
\displaystyle \bullet\ \text{Memorise all standard identities thoroughly.}
\displaystyle \bullet\ \text{Use identities instead of multiplying whenever possible.}
\displaystyle \bullet\ \text{Pay careful attention to positive and negative signs.}
\displaystyle \bullet\ \text{When expanding three terms, write every square term first and then the cross-product terms.}
\displaystyle \bullet\ \text{Use }a^3+b^3=(a+b)^3-3ab(a+b)\text{ and }a^3-b^3=(a-b)^3+3ab(a-b)\text{ whenever suitable.}
\displaystyle \bullet\ \text{If }a+b+c=0,\text{ immediately use }a^3+b^3+c^3=3abc.
\displaystyle \bullet\ \text{Read the question carefully before deciding which identity to apply.}

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