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प्रश्न
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
योग
उत्तर
( a + b )3 = a3 + 3ab( a + b ) + b3
=`( 2a)^3 + (1/(2a))^3 = 3 xx cancel(2a) xx1/cancel(2a)( 2a + 1/(2a))`
= `8a^3 + 1/(8a^3) + 3 (2a + 1/(2a))`
= `8a^3 + 1/(8a^3) + 6a + 3/(2a)`
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