## Important Algebric Formulae

- Some important Algebra formulas is given below to remember and use in solving the sums of algebra in the competitive exams.
- { (a+b) }^{ 2 } = { a }^{ 2 }+{ b }^{ 2 }+2ab or { \left[ a-b \right] }^{ 2 }+4ab
- {(a-b)}2 = {a}^2 +{b}^2 – 2ab or {(a+b)}^2 -4ab
- {a}^2 – {b}^2 = (a+b) (a-b)
- {(a+b)}^3 = {a}^3 +{b}^3 +3ab(a+b) or {a}^3 +{b}^3 +3{a}^2b+ 3a{b}^2
- {(a-b)}^3 = {a}^3 -{b}^3 -3ab(a-b) or {a}^3 -{b}^3 -3{a}^2b+ 3a{b}^2
- {a}^3 + {b}^3 = (a+b)({a}^2+{b}^2-ab) or {(a+b)}^3-3ab(a+b)
- {a}^3 – {b}^3 = (a-b)({a}^2+{b}^2 + ab) or {(a-b)}^3 + 3ab(a-b)
- {(a+b+c)}^2 = {a}^2 +{b}^2 +{c}^2 + 2(ab+bc+ca)
- {(a+b+c)}^3 = {a}^3 +{b}^3 +{c}^3 + 3(a+b)+(b+c)+(c+a)
- {a}^3 +{b}^3 +{c}^3 – 3abc = ({a}^2+{b}^2+{c}^2-ab-bc-ca)
- If a+b+c = 0 then {a}^3+{b}^3+ {c}^3 = 3abc
- {a}^3+{b}^3+ {c}^3 – 3abc = \frac{(a+b+c)}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]
- if\quad x+\frac { 1 }{ { x } } =2\quad then\quad { x }^{ n }+\frac { 1 }{ { x }^{ n } } \quad =\quad 2

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