Solve the above question and post your solution. Also you can log the time you spent on solving the question. This will give you an idea of how much is your problem solving speed.
- Tushar Sinha asked 3 months ago
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https://docs.google.com/document/d/1-hi0qKL1zmAO_NZGkYwSyoUh2lyrY6jkUgNj6tW9AU4/edit?usp=sharing
Swasti’s answer-
- swasti ranmale answered 3 months ago
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perfect cubes under 28 -> 1,8,27
if we take x = 8, then y = 20
since 20 is not a perfect cube, 8 is invalid
1 and 27 satisfy both the equations
if we take x = 27, y = 1
if we take y =27, x = 1
therefore there are 2 pairs of values that satisfy both equations simultaneously
- Devank Karkihalli answered 3 months ago
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Anirudh Seetharaman’s answer:
https://drive.google.com/file/d/1bbzLZVsaZhWQSxyPBLe3Z2EmaxvATHnA/view?usp=sharing
- Anirudh Seetharaman answered 3 months ago
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Jahaan Kukreja’s answer:
https://drive.google.com/file/d/17eVbVZq3Zl2R6VCV94QCiW2dtiA686jO/view?usp=sharing
- Jahaan Kukreja answered 3 months ago
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- prad answered 3 months ago
- This was a more rigorous version of the answer
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1^3 = 1
3^3 = 27
therefore
1+3=4
1+27=28
cube root-1=1
cube root-27=3
- Navneet Sai answered 3 months ago
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