Two variables in one equation....

[greg_rack]

Homework Statement

When simplified, 1/(1-√2)^3 is written in the form a+b√2 where a and b are integers;

what is the value of b?

Relevant Equations

No equations needed

Now, I obviously started by equalling the two expressions, but ending up with a 2-variables equation which of course can't be solved...
I cannot really understand from where could I take the second equation involving a and b to be able to put the two into a system of equations.

 

[Mark44]

Homework Statement:: When simplified, 1/(1-√2)^3 is written in the form a+b√2 where a and b are integers;

what is the value of b?
Relevant Equations:: No equations needed

Now, I obviously started by equalling the two expressions, but ending up with a 2-variables equation which of course can't be solved...
I cannot really understand from where could I take the second equation involving a and b to be able to put the two into a system of equations.

The idea is not to set the original expression equal to ##a + b\sqrt 2##. It's to work with the given expression until it has the form that is requested. What you need to do is to get rid of the ##1 -\sqrt 2## factors in the denominator. You can do this by what is called rationalizing the denominator, by multiplying by the conjugate of the denominator.
For example, ##\frac 2{a + \sqrt b} = \frac{2(a - \sqrt b)}{(a + \sqrt b)(a - \sqrt b)} = \frac{2(a - \sqrt b)}{a^2 -b}##. In the last expression, the denominator no longer has any radicals in it. In the 2nd expression, the original fraction was multiplied by 1 in the form of the conjugate of ##a + \sqrt b## over itself. It's always legitimate to multiply an expression by 1.

 

[greg_rack]

The idea is not to set the original expression equal to ##a + b\sqrt 2##. It's to work with the given expression until it has the form that is requested. What you need to do is to get rid of the ##1 -\sqrt 2## factors in the denominator. You can do this by what is called rationalizing the denominator, by multiplying by the conjugate of the denominator.
For example, ##\frac 2{a + \sqrt b} = \frac{2(a - \sqrt b)}{(a + \sqrt b)(a - \sqrt b)} = \frac{2(a - \sqrt b)}{a^2 -b}##. In the last expression, the denominator no longer has any radicals in it. In the 2nd expression, the original fraction was multiplied by 1 in the form of the conjugate of ##a + \sqrt b## over itself. It's always legitimate to multiply an expression by 1.

Got it, but still have a doubt...how do I treat the cubic power at the denominator?

 

[PeroK]

Got it, but still have a doubt...how do I treat the cubic power at the denominator?

What have you done so far?

 

[greg_rack]

What have you done so far?

I have just come up with a solution... but maybe a bit unorthodox.
That third grade at the denominator confused me, I know the process of rationalizing but not if the denominator is elevated to some exponent.
By thinking on it, I managed to reach the solution, but I'm not satisfied with the process.

 

[Office_Shredder]

Why don't you just post the whole solution you constructed, and we can tell you if there's anything wrong with it.