In the expansion of (22+)6, the coefficient of 6and 3are equal.

Find the coefficient of 6in the expansion of (1−3)(22+)6.
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1 Answer

(2x²+y)⁶?

If so, the expansion would be:

(2x²)⁶+6(2x²)⁵y+15(2x²)⁴y²+20(2x²)³y³+15(2x²)²y⁴+6(2x²)y⁵+y⁶.

The coefficient of x⁶ and y³ is 20(2³)=160.

If (1-y³)(2x²+y)⁶=(1-y³)(...+20(2x²)³y³+...)=...-160x⁶y⁶-...

The coefficient of y⁶ is therefore -160.

If (1-3y)(2x²+y)⁶=(1-3y)(...+12x²y⁵+...)=...-36x²y⁶-...

The coefficient of y⁶ is therefore -36.

Your question has a lot of missing text so it’s just a guess at what the question means.

by Top Rated User (1.2m points)

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