**Proof of the Cube Root Construction**We begin with the following diagram: The diagram is constructed with the following conditions: • AB = CD = 1 • ∠ABC = 90◦and ∠CBD = 30◦ The claim is that the segment AC has length 3√2. Proof: Just to make things easier, we will use x to represent the length of AC. By the Pythagorean Theorem, we can say that the length of BC is√x2− 1. By addition, we can say the length of AD is x + 1. The Law of Sines says that if ABC is a triangle with side lengths a, b, and c opposite the angles at A, B, and C respectively, then sinA a =sinB =sinC b c Using this on the small triangle BCD, we can say: sin∠BDC BC =sin∠CBD CD Substituting all we know, this says: sin∠BDC √x2− 1 =sin30◦ 1 Square this to say sin2∠BDC = (x2− 1)/4. Now we use the law of sines on the big triangle ABD: sin∠ADB AB =∠ABD AD Substituting all we know, this says: sin∠ADB 1 =sin120◦ x + 1Square this to say sin2∠ADB = 3/(4(x + 1)2). Finally, since ∠ADB = ∠BDC, we can set these two equations together: 4(x + 1)2=x2− 1 3 2 Cross multiply and expand, and this reduces to: x4+ 2x3− 2x − 4 = 0 which factors to (x3− 2)(x + 2) = 0 3√2!! The only positive root of this equation is