# Algebraic Problem Need Solve Urgent

• Jun 18th 2008, 10:27 PM
gautam
Algebraic 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

• Jun 18th 2008, 11:06 PM
Isomorphism
Quote:

Originally Posted by gautam
I am bothered with (Rock)a problem to solve the equation

a<sup>3</sup>+b<sup>3</sup> = 91
a<sup>2</sup>+b<sup>2</sup>= 61

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

If you mean $a^3 + b^3 = 91, a^2 + b^2 = 61$, Are a and b integers? Or are they real numbers?

By trial and error you can get one set though: a = -5 and b = 6
• Jun 18th 2008, 11:17 PM
gautam
Quote:

Originally Posted by gautam
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

• Jun 19th 2008, 12:25 AM
CaptainBlack
Quote:

Originally Posted by gautam

put $u=a+b$ and $v=ab$, then the equations become:

$u^2-2v=61$

$u^3-3uv=91$

or:

$u^2=61+2v$

$u^3=91+3uv$

Dividing gives:

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

rearranging:

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

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

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

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

RonL
• Jun 19th 2008, 05:52 AM
CaptainBlack
Quote:

Originally Posted by gautam
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