• Mar 5th 2010, 12:21 AM
Haven
Show using Bombelli's Method that $\sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 3$

So I assume the form of the expressions by letting
$1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$
$2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

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

But I can't get an answer from the equations. Any help would be greatly appreciated
• Mar 5th 2010, 01:42 AM
Hello Haven
Quote:

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

So I assume the form of the expressions by letting
$1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$
$2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

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

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 $\sqrt a + b$ and $\sqrt a - b$.

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:
$a^{\frac{3}{2}} - 3ab + 3\sqrt{a}b - b^3 = 18 -\sqrt{325}$
and then subtract this from your previous equation, before dividing by 2 to get:
$3ab + b^3 = 18$
Then it's a simple matter of eliminating $a$ by substitution:
$a = b^2+1$

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

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

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

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

$\Rightarrow \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 2b$
$=3$
• Mar 5th 2010, 06:22 AM
Haven
Thanks.
Bombelli's method is for simplifying the answers given by Cardano's formulas for cubic equations.
Cubic function - Wikipedia, the free encyclopedia
• Mar 5th 2010, 06:34 AM