How to take whole cube for a clause with n terms [closed]
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Usual formula for $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$.
What is an expansion of $(a+b+c+cdots+n)^3$?
systems-of-equations education
edited Nov 23 at 15:12
Henrik
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asked Nov 23 at 15:10
puffles
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closed as off-topic by Henrik, user302797, Gibbs, Leucippus, Cesareo Nov 24 at 1:31
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Usual formula for $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$.
What is an expansion of $(a+b+c+cdots+n)^3$?
systems-of-equations education
edited Nov 23 at 15:12
Henrik
5,92982030
asked Nov 23 at 15:10
puffles
669
closed as off-topic by Henrik, user302797, Gibbs, Leucippus, Cesareo Nov 24 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user302797, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17
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up vote
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down vote
favorite
Usual formula for $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$.
What is an expansion of $(a+b+c+cdots+n)^3$?
systems-of-equations education
edited Nov 23 at 15:12
Henrik
5,92982030
asked Nov 23 at 15:10
puffles
669
Usual formula for $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$.
What is an expansion of $(a+b+c+cdots+n)^3$?
systems-of-equations education
systems-of-equations education
edited Nov 23 at 15:12
Henrik
5,92982030
asked Nov 23 at 15:10
puffles
669
edited Nov 23 at 15:12
Henrik
5,92982030
asked Nov 23 at 15:10
puffles
669
edited Nov 23 at 15:12
Henrik
5,92982030
edited Nov 23 at 15:12
Henrik
5,92982030
edited Nov 23 at 15:12
Henrik
5,92982030
5,92982030
asked Nov 23 at 15:10
puffles
669
asked Nov 23 at 15:10
puffles
669
asked Nov 23 at 15:10
puffles
669
669
closed as off-topic by Henrik, user302797, Gibbs, Leucippus, Cesareo Nov 24 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user302797, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Henrik, user302797, Gibbs, Leucippus, Cesareo Nov 24 at 1:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, user302797, Gibbs, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17
add a comment |
1
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17
1
1
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17
add a comment |
1 Answer
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Just try it with three terms. You get
$$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a++3c^2b+6abc$$
For more terms you get each one cubed once, three times one squared times another, and six times three different terms. That reflects the number of ways to make each combination out of the three factors.
answered Nov 23 at 15:14
Ross Millikan
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1 Answer
1
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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up vote
0
down vote
Just try it with three terms. You get
$$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a++3c^2b+6abc$$
For more terms you get each one cubed once, three times one squared times another, and six times three different terms. That reflects the number of ways to make each combination out of the three factors.
answered Nov 23 at 15:14
Ross Millikan
288k23195365
add a comment |
up vote
0
down vote
Just try it with three terms. You get
$$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a++3c^2b+6abc$$
For more terms you get each one cubed once, three times one squared times another, and six times three different terms. That reflects the number of ways to make each combination out of the three factors.
answered Nov 23 at 15:14
Ross Millikan
288k23195365
add a comment |
up vote
0
down vote
up vote
0
down vote
Just try it with three terms. You get
$$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a++3c^2b+6abc$$
For more terms you get each one cubed once, three times one squared times another, and six times three different terms. That reflects the number of ways to make each combination out of the three factors.
answered Nov 23 at 15:14
Ross Millikan
288k23195365
Just try it with three terms. You get
$$(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3b^2c+3b^2a+3c^2a++3c^2b+6abc$$
For more terms you get each one cubed once, three times one squared times another, and six times three different terms. That reflects the number of ways to make each combination out of the three factors.
answered Nov 23 at 15:14
Ross Millikan
288k23195365
answered Nov 23 at 15:14
Ross Millikan
288k23195365
answered Nov 23 at 15:14
Ross Millikan
288k23195365
answered Nov 23 at 15:14
Ross Millikan
288k23195365
288k23195365
add a comment |
add a comment |
1
I think you might be looking for the multinomial theorem here.
– qmd
Nov 23 at 15:17