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Algebraic Formulae

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(i)  \large \mapsto (a+b)^{2}  =   a^{2}+b^{2}+2ab \large \Rightarrow \left (x+\frac{1}{x} \right ) ^{2} = x^{2} + \frac{1}{x^{2}}+2

\large \mapsto (a-b)^{2}  =  a^{2}+b^{2}-2ab \large \Rightarrow \left (x-\frac{1}{x} \right ) ^{2} = x^{2} + \frac{1}{x^{2}}-2

(ii) \large \mapsto (a+b)^{2}  =(a-b)^{2} +4ab   \large \Rightarrow \left (x+\frac{1}{x} \right ) ^{2} = \left (x-\frac{1}{x} \right ) ^{2} + 4

\large \mapsto (a-b)^{2}  =(a+b)^{2} -4ab   \large \Rightarrow \left (x-\frac{1}{x} \right ) ^{2} = \left (x+\frac{1}{x} \right ) ^{2} – 4

\large \mapsto (a+b)^{2} + (a-b)^{2} = 2\left (a^{2}+b^{2} \right ) \large \Rightarrow\left (x+\frac{1}{x} \right ) ^{2}+ \left (x-\frac{1}{x} \right ) ^{2}= 22\left (x^{2}+\frac{1}{x^{2}} \right )

\large \mapsto (a+b)^{2} – (a-b)^{2} =4ab \large \Rightarrow \left (x+\frac{1}{x} \right ) ^{2}– \left (x-\frac{1}{x} \right ) ^{2} = 4

(iii)  a^{2}-b^{2} = \left (a+b \right )\left (a-b \right ) \large \Rightarrow x^{2}- \frac{1}{x^{2}} = \left (x+\frac{1}{x} \right ) \times \left (x-\frac{1}{x} \right )

(iv) \large \mapsto (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)

\large \mapsto (a+b+c+d)^{2} = a^{2}+b^{2}+c^{2}+d^{2}+2ab+2ac+2ad+2bc+2bd+2cd

(v) \large \mapsto (x + a)(x + b)  =  x2  +  (a + b)x  +  ab

\large \mapsto(x + a)(x + b)(x + c)  =  x3  +  (a + b + c)x2  +  (ab + bc + ca)x  +  abc

(vi) \large \mapsto ( a + b )^{3}

= a^{3} + b^{3} +  3ab ( a + b )

= a^{3} +3a^{2}b+3ab^{2}+ b^{3}

\large \Rightarrow \left (x \pm \frac{1}{x} \right ) ^{3} = x^{3} \pm \frac{1}{x^{3}}\pm 3\left (x\pm \frac{1}{x} \right )

 (vii) \large \mapsto  a^{3}+b^{3}

=     ( a + b ) ( a^{2} –  ab  +  b^{2} )

=    ( a + b )^{3}  –  3ab ( a + b )

\large \mapsto  a^{3}-b^{3}

=     ( a – b ) ( a^{2} +  ab  +  b^{2} )

=    ( a - b )^{3}  –  3ab ( a – b )

\large \Rightarrow x^{3}+ \frac{1}{x^{3}} = \left (x \pm \frac{1}{x} \right ) \left (x^{2}+\frac{1}{x^{2}} \mp 1 \right )

= \left (x \pm \frac{1}{x} \right )^{3} \mp 3 \left (x \pm \frac{1}{x} \right )

(viii)  a2 + b2 + c2 – ab – bc – ca    =  \frac{1}{2}  [  ( a – b ) 2 +  ( b – c ) 2 + ( c – a ) 2 ]

 (ix) \large \mapsto a3 +  b3 +  c3 –  3 abc  =    ( a + b + c ) ( a2 + b2 + c2 – ab – bc – ca )

=  \frac{1}{2} ( a + b + c ) [(a – b) 2 + (b – c) 2 + (c – a) 2 ]

\large \mapsto In  above formula,

if   a + b + c  =  0  \large \Rightarrow  a3 +  b3 +  c3 = 3 abc

 

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