Answer:
d^3 - 12d^2 b + 48db^2 - 64b^3
Step-by-step explanation:
(d - 4b)^3
- First term: [3!/(3 - 0!)*0!] = 3!/(3!)*1 = 1
- First term: d^(3 - 0)*(4b)^0
- First term: d^3
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- Second Term 3!/(3 -1 )!*1! * d^(3-1)*(-4b)^1
- Second term 3 * d^2*(-4b)
- Second term - 12d^2 * b
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- Third term 3!/[(3 - 2)!2!] * d^(3 - 2) (-4b)^2
- Third term 3*d*16 b^2
- Third term 48 db^2
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- Last term 3!/[(3 - 3)!(3!) ] * d^(3 - 3) (- 4b)^(3)
- Last term 1*d^0*(-64b^3)
- Last term - 64b^3
Explanation
The general term of the binomial expansion is
<em><u>Combination</u></em>
[n! / (n - k)! k! ]
This means for the second term that k = 2 and n = 3
So the expansion becomes
3!/(3 - 1)!*1!
3!/1! 2!
3
<em><u>Descending powers</u></em>
In the second term
k = 1 ; n = 3
d^(n - k)*(-4b)^k
d^(3-1 ) * (- 4b)^(1)
d^2 * (-4b)^1
- 4 d^2 * b
Of course you have to put this together with the three
- 12 d^2 * b^1 or - 12d^2b
I know this looks awful (and it really is) but if you want the 5th term of (a - 2b)^10, you would go crazy trying to do this by expanding the binomial 10 times. So practice every one of these. Eventually it becomes mechanical. Mr. Miagi (karate kid) would say "Teacher teach. Pupil do."
