Can you Simplify Oxford University Entrance Interview ?

[@venugopalramachandran6766]

Take the conjugate as sqrt5 + sqrt 3 - sqrt8 to get 2*sqrt 15 in denominator which cancels with 30 to give sqrt 15 in numerator to be multiplied with the conjugate to give answer in fewer steps

[@superacademy247]

Thanks for sharing your alternative approach! 🔥🔥🔥I’m glad you found a more efficient way to solve the problem! 💯🙏

[@RyanLewis-Johnson-wq6xs]

Input
30/(sqrt(8) + sqrt(5) + sqrt(3)) = 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30)=5Sqrt[3]+3Sqrt[5]-2Sqrt[30]
Result
True
Logarithmic form
log(2 sqrt(2) + sqrt(3) + sqrt(5), 30) - log(2 sqrt(2) + sqrt(3) + sqrt(5), sqrt(8) + sqrt(5) + sqrt(3)) = log(2 sqrt(2) + sqrt(3) + sqrt(5), 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30))

[@rstrelba]

5 digits against 7 but simplified 🙂

[@jorgevilas1603]

Nice!

[@superacademy247]

Thanks for watching! 🙏😊I’m glad you found it interesting! 💯💖

[@RealQinnMalloryu4]

3^10/2^3+2^3+3 1^2^5/1^1^1+2^1+3 1^1/2^1+3 1/2^1+3 /2+3 (x ➖ 3x+2) .

[@tuncaytecer9062]

7,5 my answer

[@SanePerson1]

I certainly hope that this kind of pointless exercise isn't really something Oxford University is using on their entrance exams.