Expand(1/x+y/3)3. please give answer fast
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(1/x +y/3)³
= (1/x)³ + (y/3)³ + 3(1/X)²(y/3) + 3(1/x)(y/3)²
∵(a+b)³ = a³+b³+3a²b+3ab²
1/x³+y³/27 + 3(1/x²)(y/3) + 3(1/x)(y²/9)
1/x³+y³/27 + (3* y/x² * 3) + (3* y²/x * 9) canceling 3
1/x³+y³/27 + y/x² + y²/x * 3
1/x³+y³/27 + y/x² + y²/3x
= (1/x)³ + (y/3)³ + 3(1/X)²(y/3) + 3(1/x)(y/3)²
∵(a+b)³ = a³+b³+3a²b+3ab²
1/x³+y³/27 + 3(1/x²)(y/3) + 3(1/x)(y²/9)
1/x³+y³/27 + (3* y/x² * 3) + (3* y²/x * 9) canceling 3
1/x³+y³/27 + y/x² + y²/x * 3
1/x³+y³/27 + y/x² + y²/3x
Jalkaran:
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