# Solve math equation without assumptions

One mathematical sum is harassing me since a few days the problem is;:-
a^2+b^2=25 and a^3+b^3=91. find the value of a and b.
Now here we cannot take that into consoderation that as a^2+b^2=25 , a=3 and b=4. But we have to find the answer without any assumtions.May i please get the answer as soon as possible

dextercioby
Homework Helper
In which set of numbers do you search the solution to your system of equations...?

Daniel.

P.S. This is NOT "Advanced Physics".

a^3 + b^3 = (a+b)(25-ab)
Put a + b = x.
Then ab = [(a+b)^2 - (a^2 + b^2)]/2 = (x^2 - 25)/2

dextercioby
Homework Helper
It doesn't really help. A solution is to reduce everything to a 6-th order algebraic equation (either in "a" or in "b") to which you know that 3 and 4 are solutions. You can reduce the degree to 4 then and to a 4-th order algebraic equation, the exact solutions do exist. All 4 of them are complex (with nonzero imaginary part)...

That's why i asked about the set in which you want to determine the solutions...

Daniel.

Physics Monkey
Homework Helper
Just guess! There are only a few numbers that a and b even have a chance of being. In fact, there is really only one pair of numbers that could work.

VietDao29
Homework Helper
Milind_shyani said:
One mathematical sum is harassing me since a few days the problem is;:-
a^2+b^2=25 and a^3+b^3=91. find the value of a and b.
Now here we cannot take that into consoderation that as a^2+b^2=25 , a=3 and b=4. But we have to find the answer without any assumtions.May i please get the answer as soon as possible
Why do I hate people who double post so much!!! :grumpy: :grumpy: :grumpy:
Why must you post one problem in two different threads in only one board??? Why???
Maths factorization problem!!!
dextercioby said:
It doesn't really help. A solution is to reduce everything to a 6-th order algebraic equation (either in "a" or in "b") to which you know that 3 and 4 are solutions. You can reduce the degree to 4 then and to a 4-th order algebraic equation, the exact solutions do exist. All 4 of them are complex (with nonzero imaginary part)...

That's why i asked about the set in which you want to determine the solutions...

Daniel.
In fact, it can be reduced to a cubic equation. :)

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