prove; (2a+3b)^2+ (2a-3b)^2 =8a^2+18b^2
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⇔ ( 2a + 3b )^2 + ( 2a - 3b )^2
Using identities : ( a + b )^2 = a^2 + b^2 + 2ab
( a - b )^2 = a^2 + b^2 - 2ab
⇒ ( 2a + 3b )^2 + ( 2a - 3b )^2
⇒ { ( 2a )^2 + ( 3b )^2 + 2( 2a × 3b ) } + { ( 2a )^2 + ( 3b )^2 - 2( 2a × 3b ) }
⇒ { 4a^2 + 9b^2 + 12ab } + { 4a^2 + 9b^2 - 12ab }
⇒ 4a^2 + 9b^2 + 12ab + 4a^2 + 9b^2 - 12ab
⇒ 4a^2 + 4a^2 + 9b^2 + 9b^2 + 12ab - 12ab
⇒ 8a^2 + 18b^2
Hence, proved that ( 2a+3b )^2+ ( 2a-3b )^2 =8a^2 + 18b^2
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