a3+b3+c3=33 then find a,b,c?
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1
a3 + b3 + c3 = 33
Lets pick numbers that might work. This is called a trial and error. We need to pick cubic numbers that add up to 33. the number zero will not give us any results. So the three numbers must be nonzero.
1 + 8 + 27 = 36
This sum is close to 36.
-2 + 8 + 27 = 33
3√(-2)3 + 23 + 33 = 33
Therefore,
a = 3√(-2) = -1.2299
b = 2
c = 3
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Answer:a=3 ,b=4 ,c=5
Explanation:
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