NotThatSmart

Now, let's me explain my tactic. Hang with me, it's quite tedious. 

Now first off, let's set a multivariable function to deal with this question, 

First, let's let $f(a,b,c) = (ab + ac + bc)^3 - a^3 b^3 - a^3 c^3 - b^3 c^3$. (This will come into handy later)

This following step is not necessary, as you can just easily follow through, but expanding everything, which is quite tedious, we get

$3a^3b^2c+3a^2b^3c+3a^3bc^2+3a^2bc^3+3ab^3c^2+3ab^2c^3+6a^2b^2c^2$

Now, let's note that every term is divisble by 3abc. Factoring this out, we get

$f(a, b, c,)=3abc(b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2)$

The following steps are complicated, but I do believe this was explained in an AOPS course, which is where this problem came from.

(I know because I had the same problem while taking the course myself, it was a writing problem)

Let's note the following ideas. 

If we let $c=-a$, we find that $f(a,b,-a) =0$, meaning that a+c is a factor of $f(a,b,c)$ 

If we let $-a = b$, we also find that $f(a, -a, c) = 0$. This also means that a+b is a factor. 

If we let $b = -c$,  $f(a, -c, c) = 0$. This also means that b+c is a factor. 

This also means that $(a+b)(b+c)(c+a)$ should be in the final factorization. 

Wait a sec! Notice something really quickly!

$(a+b)(b+c)(c+a) = b^2c+a^2c+a^2b+ab^2+2abc+bc^2+ac^2$, which means that we can replace what we had in parethensis in the original facorization with $(a+b)(b+c)(c+a)$

Thus, our final factorization is $3abc(a+b)(b+c)(c+a)$

Thus, our final answer is $3abc(a+b)(b+c)(c+a)$

Thanks! :)

Let's graph (yay!)

What's interesting about this is this is on a coordinate plane, so we can make quite a few deductions from it. 

First off, we have that

$CD = BD = 10 \\ ED = ED$

Also though, note that ED and BC are just the x and y axis, thus they are perpendicular lines. 

Which means that we have $\angle CDE = \angle BDE$

By SAS Congruence, we know that triangle CDE is congruent to triangle BDE.

Thus, we can conclude that corresponding sides $CE = BE$

Finally, we just simply right that

$sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35$

Thus, 10.35 SHOULD be correct, not sure, trig is never my thing :(

Thanks! :)

We know that AD bisects BC. Thus, we can set variables to get some common ground here. 

Let's let $BD  = 3 - x$ and let $CD =  x$

There's a couple things we can do, but let's the Angle bisector thereom to get

$BD / AB = CD / AC$

We already set some variables, so subsituting in what we know, we get

$(3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD$

Alright, so now we know what CD is. 

Since the area of ABC can be found with CD and AB, we simply have

$[ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3$

Thus, 10/3 is our final answer. 

Thanks! :)

*1900

Let's create a system of equations to solve this question. 

First, we know that the two points must be on the equation $x^2+y^2=25$

We also know that x must be 4. 

Thus, plugging in x=4 into the first equation, we can find y values. We have

$4^2+y^2=25\\ y^2=25-16\\ y=\pm\sqrt9\\ y=3, y=-3$

Thus, the two points A and B are $(4, 3), (4, -3)$

Since they share an x value, the difference between the y values is the distance. We have

$3-(-3)=6$

Thus, 6 is our final answer. 

Thanks! :)

Let's see. The roots of a polynomial means that when we plug in x, we essentially get p(x) = 0. 

Thus, setting the polynomial to 0, we get the equation $4x^3-11x^2+2x+3 = 0$

Now, we have to find factors. We could do this in two ways. We could first have

$-4x^{3}+3x^{2}+8x^{2}-2x-3=0\\ -4x^{3}+3x^{2}+8x^{2}-6x+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+8x^{2}-6x+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+2x\cdot \left(4x-3\right)+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+2x\cdot \left(4x-3\right)+1\left(4x-3\right)=0\\-\left(4x-3\right)\cdot \left(x^{2}-2x-1\right)=0\\\left(4x-3\right)\cdot \left(x^{2}-2x-1\right)=0$

We could also test for factors. If we plug in 1, we get that $P(1) = -2$ and plugging in 0 gets $p(0) = 3$

Thus, the number must be in between 0 and 1. Testing 1/2, we get $P(1/2)=1.75$ so it must be in between 1/2 and 1. That's how we get 3/4. 

We can split these into two equations to find x. We have

$4x-3=0\\ x^{2}-2x-1=0$

Now, the second quadratic is now factorable over whole numbers, so we must use the quadratic equation. 

Using it, we get that 

$x=1+\sqrt{2}\\ x=1-\sqrt{2}$

From the first equation, we know x = 3/4 as well. 

Thus, our final 3 answers are

$x=\frac{3}{4}\\ x=1+\sqrt{2}\\ x=1-\sqrt{2}$

Thanks! :)

Let's first take a look at the fraction 1/17. 

It has 16 repeating digits in the form of $0. \overline{0588235294117647}$

We simplfy have to find which digit the 4000th lies on. 

First, let's do 4000/16. 

We get $4000/16=250$

So the 4000th digit is the last digit of the repeating decimal, which is 7. 

So 7 is our answer. 

Thanks! :)