Question

expand (y-3)wholecube​

Answer 1

abhi178

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

$\mathbb{HERE\:IS\:YOUR\:ANSWER\::-}$


${(y - 3)}^{3} \\ {y}^{3} - 3( {y}^{2})(3) + 3(y)( {3})^{2} - {3}^{3} \\ {y}^{3} - 9 {y}^{2} + 27y - 9$

$\mathbb{IDENTITY\:USED\::-}$

 (a-b)³=a³-3a²b+3ab²-b³