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Help solving equation determining numeric value [closed]

-2

$begingroup$

If $x - y = 1$ and $x^2 + y^2 = 4$

How do I show that $x^3 - y^3$ is equal to $4 + xy$?

And how do I determine the numeric value of $x^3 - y^3$?

This is an assignment question. I have scoured the tutorial video and lecture notes, I have used substitution and eliminations, I have used Symbolab, Kahn Maths and dozens of other web sites to try and work this out.

Any guidance greatly appreciated.

Thanks.

algebra-precalculus functions

share|cite|improve this question

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

$endgroup$

closed as off-topic by Dietrich Burde, Lord Shark the Unknown, RRL, Shailesh, Tianlalu Apr 3 at 1:35

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Lord Shark the Unknown, RRL, Shailesh, Tianlalu

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the first bit, use the difference of two cubes formula: $$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$$ Since $x-y=1, x^2 + y^2 = 4$, then we have $$x^3 - y^3 = (1)(4 + xy) = 4 + xy$$ I'm not sure how you can find a precise value for this without more information though.
  $endgroup$
  – Eevee Trainer
  Mar 31 at 8:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surely this is not algebraic geometry.
  $endgroup$
  – Dietrich Burde
  Mar 31 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to add relevant tags but as a newbie I didn't have enough points or something so had to choose something else close. I also tried to use the MathJax formatting, but it isn't formatting, I am missing something obviously but how do I display this correctly: x^2 Thanks
  $endgroup$
  – MCloud
  Mar 31 at 9:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dollar signs! Got it.
  $endgroup$
  – MCloud
  Mar 31 at 9:37
  

  
  
  
  

add a comment | 

-2

$begingroup$

If $x - y = 1$ and $x^2 + y^2 = 4$

How do I show that $x^3 - y^3$ is equal to $4 + xy$?

And how do I determine the numeric value of $x^3 - y^3$?

This is an assignment question. I have scoured the tutorial video and lecture notes, I have used substitution and eliminations, I have used Symbolab, Kahn Maths and dozens of other web sites to try and work this out.

Any guidance greatly appreciated.

Thanks.

algebra-precalculus functions

share|cite|improve this question

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

$endgroup$

closed as off-topic by Dietrich Burde, Lord Shark the Unknown, RRL, Shailesh, Tianlalu Apr 3 at 1:35

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Dietrich Burde, Lord Shark the Unknown, RRL, Shailesh, Tianlalu

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the first bit, use the difference of two cubes formula: $$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$$ Since $x-y=1, x^2 + y^2 = 4$, then we have $$x^3 - y^3 = (1)(4 + xy) = 4 + xy$$ I'm not sure how you can find a precise value for this without more information though.
  $endgroup$
  – Eevee Trainer
  Mar 31 at 8:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surely this is not algebraic geometry.
  $endgroup$
  – Dietrich Burde
  Mar 31 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to add relevant tags but as a newbie I didn't have enough points or something so had to choose something else close. I also tried to use the MathJax formatting, but it isn't formatting, I am missing something obviously but how do I display this correctly: x^2 Thanks
  $endgroup$
  – MCloud
  Mar 31 at 9:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dollar signs! Got it.
  $endgroup$
  – MCloud
  Mar 31 at 9:37
  

  
  
  
  

add a comment | 

$begingroup$

If $x - y = 1$ and $x^2 + y^2 = 4$

How do I show that $x^3 - y^3$ is equal to $4 + xy$?

And how do I determine the numeric value of $x^3 - y^3$?

This is an assignment question. I have scoured the tutorial video and lecture notes, I have used substitution and eliminations, I have used Symbolab, Kahn Maths and dozens of other web sites to try and work this out.

Any guidance greatly appreciated.

Thanks.

algebra-precalculus functions

share|cite|improve this question

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

$endgroup$

share|cite|improve this question

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

share|cite|improve this question

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

edited Apr 3 at 0:49

Lee David Chung Lin

4,50341342

asked Mar 31 at 8:11

MCloudMCloud

61

asked Mar 31 at 8:11

MCloudMCloud

61

2 Answers
2

active

oldest

votes

2

$begingroup$

We have $$x^3-y^3=(x-y)(x^2+xy+y^2)=1cdot (4+xy)$$ since $$x-y=1$$ and $$x^2+y^2=4$$

share|cite|improve this answer

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

add a comment | 

-1

$begingroup$

We know that
$x^3-y^3=(x-y)(x^2+xy+y^2)$ and, substituting $1$ for $x-y$ we get $x^2+xy+y^2$.

We know that $$x^2+y^2=4$$ so if we add $xy$ to both sides, we get $$x^2+xy+y^2=4+xy$$
Now $x-y=1$ so if we now multiply both sides by $1,text (which is also x-y)$ , we get the original cube difference so $x^3-y^3=4+xy$.

Now, according to WolframAlpha, there are numeric values for your consideration.

share|cite|improve this answer

edited yesterday

answered Mar 31 at 9:35

poetasispoetasis

428317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MCloud I corrected some small errors in my answer including the numeric value I had pasted from another reference. If either of the two answers so far solve your problem(s), do check it as correct to give one or the other a reputation boost.
  $endgroup$
  – poetasis
  yesterday
  

  
  
  
  

add a comment | 

2

$begingroup$

We have $$x^3-y^3=(x-y)(x^2+xy+y^2)=1cdot (4+xy)$$ since $$x-y=1$$ and $$x^2+y^2=4$$

share|cite|improve this answer

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

add a comment | 

2

$begingroup$

We have $$x^3-y^3=(x-y)(x^2+xy+y^2)=1cdot (4+xy)$$ since $$x-y=1$$ and $$x^2+y^2=4$$

share|cite|improve this answer

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

add a comment | 

$begingroup$

We have $$x^3-y^3=(x-y)(x^2+xy+y^2)=1cdot (4+xy)$$ since $$x-y=1$$ and $$x^2+y^2=4$$

share|cite|improve this answer

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

share|cite|improve this answer

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Mar 31 at 8:18

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

-1

$begingroup$

We know that
$x^3-y^3=(x-y)(x^2+xy+y^2)$ and, substituting $1$ for $x-y$ we get $x^2+xy+y^2$.

We know that $$x^2+y^2=4$$ so if we add $xy$ to both sides, we get $$x^2+xy+y^2=4+xy$$
Now $x-y=1$ so if we now multiply both sides by $1,text (which is also x-y)$ , we get the original cube difference so $x^3-y^3=4+xy$.

Now, according to WolframAlpha, there are numeric values for your consideration.

share|cite|improve this answer

edited yesterday

answered Mar 31 at 9:35

poetasispoetasis

428317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MCloud I corrected some small errors in my answer including the numeric value I had pasted from another reference. If either of the two answers so far solve your problem(s), do check it as correct to give one or the other a reputation boost.
  $endgroup$
  – poetasis
  yesterday
  

  
  
  
  

add a comment | 

-1

$begingroup$

We know that
$x^3-y^3=(x-y)(x^2+xy+y^2)$ and, substituting $1$ for $x-y$ we get $x^2+xy+y^2$.

We know that $$x^2+y^2=4$$ so if we add $xy$ to both sides, we get $$x^2+xy+y^2=4+xy$$
Now $x-y=1$ so if we now multiply both sides by $1,text (which is also x-y)$ , we get the original cube difference so $x^3-y^3=4+xy$.

Now, according to WolframAlpha, there are numeric values for your consideration.

share|cite|improve this answer

edited yesterday

answered Mar 31 at 9:35

poetasispoetasis

428317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MCloud I corrected some small errors in my answer including the numeric value I had pasted from another reference. If either of the two answers so far solve your problem(s), do check it as correct to give one or the other a reputation boost.
  $endgroup$
  – poetasis
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

We know that
$x^3-y^3=(x-y)(x^2+xy+y^2)$ and, substituting $1$ for $x-y$ we get $x^2+xy+y^2$.

We know that $$x^2+y^2=4$$ so if we add $xy$ to both sides, we get $$x^2+xy+y^2=4+xy$$
Now $x-y=1$ so if we now multiply both sides by $1,text (which is also x-y)$ , we get the original cube difference so $x^3-y^3=4+xy$.

Now, according to WolframAlpha, there are numeric values for your consideration.

share|cite|improve this answer

edited yesterday

answered Mar 31 at 9:35

poetasispoetasis

428317

$endgroup$

share|cite|improve this answer

edited yesterday

answered Mar 31 at 9:35

poetasispoetasis

428317

answered Mar 31 at 9:35

poetasispoetasis

428317

answered Mar 31 at 9:35

poetasispoetasis

428317