Given x+y = 3 and xy = 7, Evaluate x^5 + y^5

The idea behind the binomial expansion solution is to avoid having to solve the given system of equations for x and y. The answer to question is attainable without EVER knowing the values of the variables x and y.
Southern Illinois University mathematician/historian
Non-Intersecting Line And Hyperbola result in the Sum of Two Fifth Power irrational imaginary part complex conjugates.
University math professor Michael W. Ecker refused to answer a question which he characterized as a "poor one" over complex numbers. He rightly eschews mere formulaic chug-and-plug, but knowing which formula(s) to use is immensely important.The question was posed by a Quora user. Part of being a good teacher is creating a learning moment from a so called bad question, rather than berating a student trying to learn who may already have low self-esteem. The cyclic number enthusiast will be better at his craft when he recognizes this truth.

[@DonRedmond-jk6hj]

A little less messy is to compute x^2 + y^2 and x^3 + y^3. Multiply the two and bingo/

[@MyOneFiftiethOfADollar]

We didn’t get a chance to see how “messy” your approach was.
Provide more detail for your brilliant idea.
Southern Illinois University mathematician/historian

[@franciscook5819]

Same idea, slightly different approach ...
(x+y)⁵=243=x⁵+5x⁴y+10x³y²+10x²y³+5xy⁴+y⁵
=x⁵+y⁵+5xy(x³+2x²y+2xy²+y³)
=x⁵+y⁵+35(x³+3x²y+3xy²+y³-(x²y+xy²))
=x⁵+y⁵+35((x+y)³-xy(x+y))
=x⁵+y⁵+35(27-21)
=x⁵+y⁵+210
x⁵+y⁵=33

[@MyOneFiftiethOfADollar]

Both approaches involved expanding x+y to the 5th and 3rd and substituting.
I like this one because x and y are complex numbers with irrational coefficients. Raising them to 5th power is messy as far as I know.
The method we used avoids having to directly solve for x and y.