As an engineer who has used math my entire career, I so enjoy your elegant solutions to all these problems.  Great job, so entertaining!

I take "shortcuts" in my mind, but I love Susanne's step by step.

Except for my geometry teacher (proofs), my teachers hated me, because I did so many steps in my head. They would've loved Susanne.

Thanks for the flashback to high school math class, now I feel old. Really enjoy your videos, I always try to solve them myself first usually get them but my solutions are rarely as elegant as yours.

I always wished I had a teacher, like Susanne, at school, who could explain how to simplify an equation step by step in such a logical and elegant way. Many people can do algebra, but being able to explain it in a way that everyone can understand is the art of teaching. Thank you, Susanne !
There is one step, though, that I would have done differently.
I would have used the term "bracketing out" (factoring out) the common term "five to the power of one third" in the fourth root of the bracket, two times five to the power of one third plus three times five to the power of one third.
I would have used the term "bracketing out" (or "factoring out") the common term "five to the power of one third", which would have resulted in the entire expression simplifying beautifully down to just "five to the power of one third".

Thank you. Once again you've shown me I have forgotten more than I remember. I love the refreshers

I'm always impressed with the process. Great video.

I did this in my head ... and got it hopelessly wrong! Along comes Susanne with her calm and methodical approach and untangled the mental knots I'd tied myself up in.

This series is the best maths tuition I have ever seen.

My mom helped me so much in school.

Thank you, Susanne. ❤❤

Great job, dear professor, thanks so much ❤❤

(40^(1/3) + 135^(1/3))^(1/4)

(2(40/2^3)^(1/3) + 3(135/3^3)^(1/3))^(1/4)

(2(5)^(1/3) + 3(5)^(1/3))^(1/4)

(5(5)^(1/3))^(1/4)

((5*5^3)^(1/3))^(1/4)

((625)^(1/3))^(1/4)

625^(1/(3*4))

5^(1/3)

Muito bom. Boa didática.

As a maths failure, I am learning something.

If R is the given radical, then:
R = ⁴√((2+3)•³√5) = (5^(1 + 1/3))^(1/4) = 5^(4/3 • 1/4) = ³√5.

Great video, ma'am!

40^(1/3) + 135^(1/3) = 2•5^(1/3) + 3•5^(1/3) = 5•5^(1/3) = [(5^3)•5]^(1/3) = [(5^4]^(1/3)
{[(5^4]^(1/3)}^(1/4) = 5^(1/3) ✅

I started by factoring into primes: 40 = 2^3 * 5 and 135 = 3^3 * 5.  After you bring the 2 and 3 out, you're left with 5^(1/3) in both terms.

Excellent!