Hey, Here's the search for the most intelligent guy!!

First of all , I want to tell that they are no tricky problems. Tooooooo easy problems.

The complete solutions are -

1.) Clearly, by Vieta's Formula - $a+b+c=1$, $ab+bc+ca = 1$, $abc=2$

We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c)((a+b+c)^2 -3(ab+bc+ca))$

Plugging the values, we get

$a^3+b^3+c^3-6=(1)(1^2-3(1))=-2 \Rightarrow a^3+b^3+c^3 = 4$

2.) Since ABC is similar to DCA, the ratio of their corresponding sides must be equal. So,

$\frac{AB}{DC}=\frac{BC}{AC} \Rightarrow DC = 12$

3.) The given equations can be simplified to $yz + 2xz + 3xy = 0$ and $yz - 6xz - 5xy = 0$

Subtracting both the equations, we get -

$z + y = 0 \Rightarrow z=-y$

Substituting $z=-y$ in first equation, we get - $x^2-y^2 = 0$

Now, put $z=-y$ in $x/y + y/z + z/x$. So, what you will get is $x/y + y/z + z/x = -1$

4.) $a+b = 1$, $a^2+b^2 = 2$

We can easily get $ab = \frac{-1}{2}$

Now -

$a^5 + b^5 = a^5 + b^5 + a^3b^2 + a^2b^3 - a^3b^2 - a^2b^3$

$= a^3(a^2+b^2)+b^3(a^2+b^2) - a^3b^2 - a^2b^3 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$

So, $a^5+b^5 = (a^2+b^2)(a^3+b^3) - (ab)^2(a+b) = (a^2+b^2)(a+b)(a^2+b^2-ab) - (ab)^2(a+b)$

Now plugging the values, we get - $a^5+b^5 = \frac{19}{4}$

5.) First, let $\angle DRC = \angle BRp = \theta$ and $RP = x$

Then $AD = (5+x)cos \theta$ and $RC = 5cos \theta$

Now observe that - triangle QAD is similar to QCR. So, on solving for PR gives $PR = \frac{5}{2}$

6.) Let $BD = x$, $\angle BAD = \angle DAC = \theta$

Then, from triangle $ABC \rightarrow sin 2\theta = \frac{8}{10} = \frac{4}{5}$

from triangle $BAD \rightarrow sin \theta = \frac{x}{\sqrt{36+x^2}}$ and $cos \theta = \frac{6}{\sqrt{36+x^2}}$

We know that $sin 2\theta = 2sin\theta cos\theta$

Plugging their values, we get the following quadratic - $s^2-15x+36 \Rightarrow x = 12, 3$

Since, $BC = 8$, so we can't take $x=12$

Hence, $AD = \sqrt{3^2 + 6^2} = \sqrt{45}$

Next time when you post a question in discussion, please makes sure that they are of proper difficulty level.

I'm not trying to be rude at all but these questions were really easy and they aren't science problems. Though if you are having trouble with anyone of them just let me know because I will help you. I can give you some hints if you like too: 1. this involves complex numbers 2. I used the cosine rule 3. this is just algebra 4. this just involves a quadratic 5. just a bit of trigonometry 6. this just involves Pythagoras/tan. Hope this helps.

These are very easier problems and way below than the level of Brilliant problems. You must not post these types of problems to this discussion room.

that was easy. Now, help me solve this... $x^{2}+5x+5=32$ Don't use trial and error method and the $x$ value should be a single one.