Solve the question..........
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i think it's enough for
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( A-B ) ^ 3
A^3 - B^3 - 3ab ( a-b )
( x - 1/x ) = b
( x - 1/x ) ^ 3 = ( b ) ^3
x^3 - 1/x^3 - ( 3*x*1/x ) ( x- 1/x ) = b^3
x^3 - 1/x^3 - 3 ( x - 1/x ) = b^3
x^3 - 1/x^3 - 3x + 3/x = b^3
x^3 - 1/x^3 = b^3 + 3x - 3/x ( ANS )
Pls mark as brainliest answer
A^3 - B^3 - 3ab ( a-b )
( x - 1/x ) = b
( x - 1/x ) ^ 3 = ( b ) ^3
x^3 - 1/x^3 - ( 3*x*1/x ) ( x- 1/x ) = b^3
x^3 - 1/x^3 - 3 ( x - 1/x ) = b^3
x^3 - 1/x^3 - 3x + 3/x = b^3
x^3 - 1/x^3 = b^3 + 3x - 3/x ( ANS )
Pls mark as brainliest answer
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