An Evil Question!

Since alpha is in cubic root , so I make the sum given in a cube.

$10 + 6\sqrt{3} = 1^3 + (\sqrt{3})^3 + 3\sqrt{3} + 9$

$10 + 6\sqrt{3}= 1 + (\sqrt{3})^3 + 3\sqrt{3}(1 + \sqrt{3})$

$10 + 6\sqrt{3} = (1 + \sqrt{3})^3$

$\alpha = 1 + \sqrt{3}$

$(1 - \sqrt{3})(1 + \sqrt{3}) = 1 - 3 = -2$

Using calculator, the answer is -2.

I cubed alpha first to get 10 + 6root3. Then I cubed 1 - root3 as well. I multiplied the cubed values which gave me -8 and then I found the cube root which gives -2.

seeing the power 1/3, i think of making 10 +6root3 into a number with 3 as the power, which is (1 +3.1^2.root3 + 3 1 (root3)^2 + 3root3)= (1+root3)^3. thus the answer is (1+r3)(1-r3)= -2