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ChiralSuperfields
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- Homework Statement
- Please see below
- Relevant Equations
- Please see below
How did they get y^7 in the bottom fraction? I got their answer except I had y^6. Would some please be able to help?
Many thanks!
Multiply both the top and bottom by ##y## to help simplify the numerator...Callumnc1 said:Homework Statement:: Please see below
Relevant Equations:: Please see below
How did they get y^7 in the bottom fraction?
Thank you for your reply @berkeman! I see now :)berkeman said:Multiply both the top and bottom by ##y## to help simplify the numerator...
The old saying is "close, but no cigar."Callumnc1 said:How did they get y^7 in the bottom fraction? I got their answer except I had y^6.
Thank you for your reply @Mark44!Mark44 said:The old saying is "close, but no cigar."
Obviously, you did something wrong. If we look just at the numerator of the fraction in (1), we have:
$$3x^2y^3 - 3x^3y^2(-\frac{x^3}{y^3}) = 3x^2y^3 + \frac{3x^6}y$$
So the fraction in (1) can be rewritten as $$\frac{3x^2y^3 + \frac{3x^6}y}{y^6}$$
What do you need to do to turn this complex fraction into an ordinary fraction; i.e., one that is the quotient of two polynomials?
That's not a good way to think about it. The next step from where I left off is to multiply the complex fraction by 1, in the form of ##\frac y y##. That will bump the exponent on y in the first term up top and the term in the denominator, and will clear the fraction in the second term up top. You can always multiply by 1 without changing the value of the thing that is being multiplied.Callumnc1 said:You multiply the ##3x^2y^3## and ##\frac{3x^6}{y}## by ##y## then flip the ##y^6## up
Thank you for your reply @Mark44!Mark44 said:That's not a good way to think about it. The next step from where I left off is to multiply the complex fraction by 1, in the form of ##\frac y y##. That will bump the exponent on y in the first term up top and the term in the denominator, and will clear the fraction in the second term up top. You can always multiply by 1 without changing the value of the thing that is being multiplied.
Missing out on concepts like this is why I recommended spending some time going over basic precalculus topics in another thread you posted.