Question #69c6c

Question

1498 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-08T16:41:27

I get $-200$

Expand $(x-2/x)^6 = (x-2x^-1)^6$ to get

$x^6 - 6x^5(2x^-1)+15x^4(4x^-2)-20x^3(8x^-3)+15x^3(16x^-4) - 6x^2(32x^-5)+(64x^-6)$

When we distribute the $2$ from $(2+3/x^2)$ the product with every term except $-20x^3(8x^-3)$ will depend on $x$ . That product is

$2 xx -20 xx 8 = -320$

And when we distribute the $3/x^2$ the product of $3/x^2$ and $15x^4(4x^-2)$ will be independent of $x$ . That product is

$2 xx 15 xx 4 = 120$

The sum of those two products is

$-200$