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Binomial Expansion based on equation for evaluation
Applying the binomial expansion to a sum of multiple binomial expansions.Help with binomial expansion exerciseConvergence of a binomial expansionBinomial Expansion - Finding the term independent of n.In the expansion of $(1+x+x^2)^n$, find the required valueApproximating square roots using binomial expansion.Coefficient in binomial expansion for negative termsBinomial Series ExpansionCoefficient of $x$ in a finite expansionBinomial Expansion confirmation
$begingroup$
I have this question that has really stumped me, it is supposed to be done via Binomial kind of expansion.
If $x+frac1x=10$ find the value of $x^3+frac1{x^3}$.
So I hope some one has an approach to this question.
algebra-precalculus binomial-coefficients binomial-theorem
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
$endgroup$
add a comment |
$begingroup$
I have this question that has really stumped me, it is supposed to be done via Binomial kind of expansion.
If $x+frac1x=10$ find the value of $x^3+frac1{x^3}$.
So I hope some one has an approach to this question.
algebra-precalculus binomial-coefficients binomial-theorem
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
$endgroup$
$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43
add a comment |
$begingroup$
I have this question that has really stumped me, it is supposed to be done via Binomial kind of expansion.
If $x+frac1x=10$ find the value of $x^3+frac1{x^3}$.
So I hope some one has an approach to this question.
algebra-precalculus binomial-coefficients binomial-theorem
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
$endgroup$
I have this question that has really stumped me, it is supposed to be done via Binomial kind of expansion.
If $x+frac1x=10$ find the value of $x^3+frac1{x^3}$.
So I hope some one has an approach to this question.
algebra-precalculus binomial-coefficients binomial-theorem
algebra-precalculus binomial-coefficients binomial-theorem
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
edited Mar 11 at 3:17
Parcly Taxel
44.5k1376109
44.5k1376109
asked Mar 10 at 22:36
PaluPalu
3582823
asked Mar 10 at 22:36
PaluPalu
3582823
asked Mar 10 at 22:36
PaluPalu
3582823
3582823
$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43
add a comment |
$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43
$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that
$$
left(x+frac{1}{x}right)^3=x^3+frac{1}{x^3}+3x^2left(
frac{1}{x}
right)+3xleft(
frac{1}{x^2}
right)=x^3+frac{1}{x^3}+3left(x+frac{1}{x}right)
$$
You are given enough information to solve for the required quantity.
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
add a comment |
$begingroup$
Hint:
Expand $;biggl(x+dfrac1xbiggr)^3$ by the standard formula.
answered Mar 10 at 22:41
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
Not sure how to do this with a Binomial expansion but:
$x+frac1x=10 to x^2+1=10x to x^2-10x+1=0$
Solve for $x$ using quadratic formula and plug it into $x^3+frac1{x^3}$.
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$
left(x+frac{1}{x}right)^3=x^3+frac{1}{x^3}+3x^2left(
frac{1}{x}
right)+3xleft(
frac{1}{x^2}
right)=x^3+frac{1}{x^3}+3left(x+frac{1}{x}right)
$$
You are given enough information to solve for the required quantity.
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
add a comment |
$begingroup$
Note that
$$
left(x+frac{1}{x}right)^3=x^3+frac{1}{x^3}+3x^2left(
frac{1}{x}
right)+3xleft(
frac{1}{x^2}
right)=x^3+frac{1}{x^3}+3left(x+frac{1}{x}right)
$$
You are given enough information to solve for the required quantity.
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
add a comment |
$begingroup$
Note that
$$
left(x+frac{1}{x}right)^3=x^3+frac{1}{x^3}+3x^2left(
frac{1}{x}
right)+3xleft(
frac{1}{x^2}
right)=x^3+frac{1}{x^3}+3left(x+frac{1}{x}right)
$$
You are given enough information to solve for the required quantity.
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
$endgroup$
Note that
$$
left(x+frac{1}{x}right)^3=x^3+frac{1}{x^3}+3x^2left(
frac{1}{x}
right)+3xleft(
frac{1}{x^2}
right)=x^3+frac{1}{x^3}+3left(x+frac{1}{x}right)
$$
You are given enough information to solve for the required quantity.
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
answered Mar 10 at 22:49
Foobaz JohnFoobaz John
22.7k41452
22.7k41452
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
add a comment |
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
Thanks Foobaz, i did this same expansion and got what you got, and somehow could not see the repetitive structure of (x+1/x) somehow, which has a value of 10. I can now clearly see the solution. I guess it was one of those days where one can't see the obvious.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
$begingroup$
I will choose your post as the Answer to this question. Thanks once more Foobaz.
$endgroup$
– Palu
Mar 11 at 1:45
add a comment |
$begingroup$
Hint:
Expand $;biggl(x+dfrac1xbiggr)^3$ by the standard formula.
answered Mar 10 at 22:41
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
Hint:
Expand $;biggl(x+dfrac1xbiggr)^3$ by the standard formula.
answered Mar 10 at 22:41
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
Hint:
Expand $;biggl(x+dfrac1xbiggr)^3$ by the standard formula.
answered Mar 10 at 22:41
BernardBernard
123k741116
$endgroup$
Hint:
Expand $;biggl(x+dfrac1xbiggr)^3$ by the standard formula.
answered Mar 10 at 22:41
BernardBernard
123k741116
answered Mar 10 at 22:41
BernardBernard
123k741116
answered Mar 10 at 22:41
BernardBernard
123k741116
answered Mar 10 at 22:41
BernardBernard
123k741116
123k741116
add a comment |
add a comment |
$begingroup$
Not sure how to do this with a Binomial expansion but:
$x+frac1x=10 to x^2+1=10x to x^2-10x+1=0$
Solve for $x$ using quadratic formula and plug it into $x^3+frac1{x^3}$.
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
$endgroup$
add a comment |
$begingroup$
Not sure how to do this with a Binomial expansion but:
$x+frac1x=10 to x^2+1=10x to x^2-10x+1=0$
Solve for $x$ using quadratic formula and plug it into $x^3+frac1{x^3}$.
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
$endgroup$
add a comment |
$begingroup$
Not sure how to do this with a Binomial expansion but:
$x+frac1x=10 to x^2+1=10x to x^2-10x+1=0$
Solve for $x$ using quadratic formula and plug it into $x^3+frac1{x^3}$.
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
$endgroup$
Not sure how to do this with a Binomial expansion but:
$x+frac1x=10 to x^2+1=10x to x^2-10x+1=0$
Solve for $x$ using quadratic formula and plug it into $x^3+frac1{x^3}$.
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
answered Mar 10 at 22:45
Akash PatelAkash Patel
15919
15919
add a comment |
add a comment |
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$begingroup$
What have you tried? What is the closest thing here that looks like a binomial?
$endgroup$
– darij grinberg
Mar 10 at 22:38
$begingroup$
Hi Darji, just like below where one tried to expand out the x +1/x all raised to the power of 3, i got that expansion, but could not see it further(it was to much for me to write it out here). But Foobaz below did the same as i did, but after he did it, i can see how simple the solution was.
$endgroup$
– Palu
Mar 11 at 1:43