Simplifying this complicated fraction

[ChiralSuperfields]

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!

 

[berkeman]

Homework Statement:: Please see below
Relevant Equations:: Please see below

How did they get y^7 in the bottom fraction?

Multiply both the top and bottom by $y$ to help simplify the numerator...

 

[ChiralSuperfields]

Multiply both the top and bottom by $y$ to help simplify the numerator...

Thank you for your reply @berkeman! I see now :)

 

[Mark44]

How did they get y^7 in the bottom fraction? I got their answer except I had y^6.

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?

 

[ChiralSuperfields]

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?

Thank you for your reply @Mark44!

You multiply the $3x^2y^3$ and $\frac{3x^6}{y}$ by $y$ then flip the $y^6$ up

Many thanks!

 

[Mark44]

You multiply the $3x^2y^3$ and $\frac{3x^6}{y}$ by $y$ then flip the $y^6$ up

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.

 

[ChiralSuperfields]

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.

Thank you for your reply @Mark44!

That is a very good method you mention! My teacher taught be that method that I was talking about. Your method makes sense :)