I am bothered with (Rock)a problem to solve the equation

a3+b3 = 91

a2+b2=61

find the value of and "a" and "b".

where 3 and two is in form of cube and square

Please help me to solve this problem(Smirk)

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- Jun 18th 2008, 09:27 PMgautamAlgebraic Problem Need Solve Urgent
I am bothered with (Rock)a problem to solve the equation

a3+b3 = 91

a2+b2=61

find the value of and "a" and "b".

where 3 and two is in form of cube and square

Please help me to solve this problem(Smirk) - Jun 18th 2008, 10:06 PMIsomorphism
- Jun 18th 2008, 10:17 PMgautam
- Jun 18th 2008, 11:25 PMCaptainBlack
put $\displaystyle u=a+b$ and $\displaystyle v=ab$, then the equations become:

$\displaystyle u^2-2v=61$

$\displaystyle u^3-3uv=91$

or:

$\displaystyle u^2=61+2v$

$\displaystyle u^3=91+3uv$

Dividing gives:

$\displaystyle u=\frac{91+3uv}{61+2v}$

rearranging:

$\displaystyle

v=\frac{61u-91}{u}

$

Now substitute this back into $\displaystyle u^2-2v=61$ to get:

$\displaystyle u^2-2\frac{61u-91}{u}=61$

Which is a cubic in $\displaystyle u$, solve this, then use these solutions to find the corresponding $\displaystyle v$'s and so the $\displaystyle a$'s and $\displaystyle b$'s. (solution of the cubic will be eased by using Isomorphism's result).

RonL - Jun 19th 2008, 04:52 AMCaptainBlack
- Jun 19th 2008, 06:24 AMcolby2152
I just wanted to mention that those two solutions posted by Captain Black were excellent.(Rock)