Hello Haven Originally Posted by

**Haven** Show using Bombelli's Method that $\displaystyle \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 3$

So I assume the form of the expressions by letting

$\displaystyle 1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$

$\displaystyle 2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

Multiply 1 and 2 and you get: $\displaystyle 1 = a - b^2$

Cube both sides of 1 and you get: $\displaystyle a^{\frac{3}{2}} + 3ab + 3\sqrt{a}b + b^3 = 18 + \sqrt{325}$

Which means $\displaystyle a^{\frac{3}{2}} + 3\sqrt{a}{b} = \sqrt{325}$

and $\displaystyle 3ab + b^3 = 18$

But I can't get an answer from the equations. Any help would be greatly appreciated

Bombelli's method is new to me! I should not have thought of using $\displaystyle \sqrt a + b$ and $\displaystyle \sqrt a - b$.

I agree with your equations, although I think that in order to justify the last two you also need to cube both sides of (2) to get:

$\displaystyle a^{\frac{3}{2}} - 3ab + 3\sqrt{a}b - b^3 = 18 -\sqrt{325}$

and then subtract this from your previous equation, before dividing by 2 to get:

$\displaystyle 3ab + b^3 = 18$

Then it's a simple matter of eliminating $\displaystyle a$ by substitution:

$\displaystyle a = b^2+1$

$\displaystyle \Rightarrow 3(b^2+1)b + b^3 = 188$

$\displaystyle \Rightarrow 4b^3+3b-18=0$

$\displaystyle \Rightarrow (2b-3)(2b^2-3b+6)=0$

$\displaystyle \Rightarrow b = \tfrac32$, since the discriminant of the second factor is negative

$\displaystyle \Rightarrow \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 2b$

$\displaystyle =3$

Grandad