How do you expand #(d-1/d)^5#?

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Answer 1 · 2018-04-27T12:54:12

Two ways to do it. Both using Newton's binomial theorem

Way 1 .

Apply directly Newton's theorem (alternate + and -)

#(d-1/d)^5=5C0d^5-5C1d^4·1/d^1+5C2d^3·1/d^2-5C3d^2·1/d^3+5C4d^1·1/d^4-5C5d^0·1/d^5# where

#mCn# are terms of Pascal triangle in the 5th row which are

5C0=1, 5C1=5, 5C2=10, 5C3=10, 5C4=5 and 5C5=1

And result in #d^5-5d^3+10d-10/d-5/d^3-1/d^5#

Way 2

Operate in #(d-1/d)^5=((d^2-1)/d)^5# and apply Newton's Theorem in numerator and denominator (which is simple #d^5#)