Question

Using binomial theorm, expand (2+x+x²)³

Answer 1

(2+x(1+X))^3= 3!/0!*(x(1+X))3*2^0+3!/(2!*1!)*(x(1+x))^2*2^1 +3!/(2!*1!)*(x(1+x))^1*2^2+

3!(3!*0!)*9x(1+x))^0*2^3.......

now we can expand as usual.....

Answer 2

(1 - 2.x)³.(1 - 2/x)³ = [(1 - 2.x).(1 - 2/x)]³ = [1 - 2/x - 2.x + 2.x.(2/x)]³ = [5 - 2.(x + 1/x)]³

When we expand the last expression using the binomial theorem, we note that there are only two terms which can produce a constant value - the initial 5³ and that containing (x + 1/x)². Those with (x + 1/x) raised to an odd power can only produce terms in x^n and x^(-n), where n ranges from 1 to 3. To illustrate the point I'll write the expansion out in full

[5 - 2(x + 1/x)]³ = 5³ - 3*5².2.(x - 1/x) + 3*5*2².(x + 1/x)² - 2³.(x + 1/x)³

With (x + 1/x)² = [x² + 2 + (1/x)²], the only constant terms are given by 5³ + 3*5*2².(2)

= 125 + 120 = 245.

And, yes, in view of the spread of results, I checked the answer using the brute force method long-hand and with WolframAlpha. These gave

(1 - 2.x)³.(1 - 2/x)³ = [1 - 6.x + 12.x² - 8.x³].[1 - 6/x + 12/x² - 8/x³]

= 1 + 6² + 12² + 8² = 245 for the constant term. 1 Comment
Other Answers (3) Oldest walter answered 2 years ago The work would be a mess and isn't needed to find the answer (in fact, that really the point of this question).

A bunch of stuff is going to be multiplied together, but the only product that won't have an x somewhere in it will be 1*1 (the first term in each of the two parenthesis).

So your constant term will be 1.

Everything else will have either 2x or 2/x (or both) as a factor, so they won't be constants.