Question

(1/x+y/3)whole cube

Answer 1

Answer:

we have to expand (1/x + y/3)³

we know,

(a + b)³ = (a + b)²(a + b)

= (a² + 2ab + b²)(a + b)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + 3a²b + 3ab² + b³

so, (1/x + y/3)³ = (1/x)³ + 3(1/x)²(y/3) + 3(1/x)(y/3)² + (y/3)³

= 1/x³ + y/x² + y²/3x + y³/27

hence, (1/x + y/3)³ = 1/x³ + y/x² + y²/3x + y³/27

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Answer 2

Answer:

we have to expand (1/x + y/3)³

we know,

(a + b)³ = (a + b)²(a + b)

= (a² + 2ab + b²)(a + b)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + 3a²b + 3ab² + b³

so, (1/x + y/3)³ = (1/x)³ + 3(1/x)²(y/3) + 3(1/x)(y/3)² + (y/3)³

= 1/x³ + y/x² + y²/3x + y³/27

hence, (1/x + y/3)³ = 1/x³ + y/x² + y²/3x + y³/27

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