How do you use the binomial series to expand #(2x^2 – (3/x) ) ^8#?

1 Answer

Answer:

#(2x^2-3/x)^8=256x^16-3072x^13+16128x^10,-48384x^7+90720x^4-108864x+81648x^(-2)-34992x^(-5)+6561x^-8#

Explanation:

Binomial Expansion follows the following general rule:

#(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n#

In this question, #a=2x^2, b=-3/x#

I'll use a chart to find each term:

#((C, a, b, term),(1,256x^16,1,256x^16),(8,128x^14,-3/x,-3072x^13),(28,64x^12,9/x^2,16128x^10),(56,32x^10,-27/x^3,-48384x^7),(70,16x^8,81/x^4,90720x^4),(56,8x^6,-243/x^5,-108864x),(28,4x^4,729/x^6,81648x^(-2)),(8,2x^2,-2187/x^7,-34992x^(-5)),(1,1,6561/x^8,6561x^-8))#

The final answer is the sum of all the terms:

#(2x^2-3/x)^8=256x^16-3072x^13+16128x^10-48384x^7+90720x^4-108864x+81648x^(-2)-34992x^(-5)+6561x^-8#