Gubbubu

Alkalmazva az <math> A^{2} - B^{2} = \left( A+B \right) \left( A-B \right) </math> „nevezetes azonosságot” az <math> A = 2bc </math> és <math>B = -a^{2}+b^{2}+c^{2} </math> esetekre;
 
:: <math>X</math> &nbsp; <math> = </math> &nbsp; <math> \frac { \left[ \left( 2bc \right) + \left( -a^{2}+b^{2}+c^{2} \right) \right] \left[ \left( 2bc \right) - \left( -a^{2}+b^{2}+c^{2} \right) \right] } { \left( 2c \right) ^{2} } </math> &nbsp; <math> = </math> &nbsp; <math> \frac { \left( -a^{2}+b^{2}+2bc+c^{2} \right) \left( 2bc + a^{2}-b^{2}-c^{2} \right) } { \left( 2c \right) ^{2} } </math> &nbsp; <math> = </math> <br> <math> = </math> &nbsp; <math> \frac { \left( -a^{2}+b^{2}+2bc+c^{2} \right) \left( a^{2}-b^{2}+2bc-c^{2} \right) } { \left( 2c \right) ^{2} } </math> &nbsp; <math> = </math> <math> \frac { \left[ -a^{2} + \left( b+c \right)^{2} \right] \left[ a^{2}- \left( b+c \right) ^{2} \right] } { \left( 2c \right) ^{2} } </math> &nbsp; <math> = </math> &nbsp; <math> \frac { \left[ -a^{2} + \left( b+c \right)^{2} \right] \left[ a^{2}- \left( b+c \right) ^{2} \right] } { \left( 2c \right) ^{2} } </math>