2015/106) Factorize: 2x^2-5xy-3y^2+3x+19y-20

The factor is of the form $(ax+by+c)(dx+ey+f)$

if we ignore c and f and multiply we get

$(ax+by)(dx+ey)= adx^2 + (ae+bd)xy + bey^2$

so we can first factor $2x^2-5x-3y^2$ and then evaluate the other parts

$2x^2-5xy-3y^2= (2x+y)(x-3y)$ factored by quadratic method

so

 $2x^2-5xy-3y^2+3x+19y-20= (2x+y+e)(x-3y+f)$

      $= 2x^2-3y^2-5xy+(2f+e)x + (f-3e)y + ef$

Comparing coefficients of x , y and constant separately we get
$2f+e = 3, f- 3e = 19$ and $ef = - 20$
solving 1^st 2 equations we get $e = -5$ and $f = 3$ and 3^rd equation also meets criteria
so we get 

$2x^2-5xy-3y^2+3x+19y-20= (2x+y-5)(x-3y+4)$

2015/105) if $a=xy^2+yz^2+zx^2$ and $b=x^2y+y^2z+z^2x$ then express $(x^3-y^3)(y^3-z^3)(z^3-x^3)$ in terms of a and b

$(x^3-y^3)(y^3-z^3)(z^3-x^3)$
= $(x^3-y^3)(y^3z^3-y^3x^3-z^6+z^3x^3)$
=$x^3y^3z^3-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6-x^3z^3x^3$
=  $-y^3x^6-z^6x^3+z^3x^6-y^6z^3+y^6x^3+y^3z^6$
= $(z^3x^6+y^3 z^6 + x^3y^6) - (y^3x^6 + x^3z^6 + y^6z^3)$
HENCE
$(x^3-y^3)(y^3-z^3)^(z^3-x^3)$
  $= ((xy^2)^3 + (yz^2)^3+(zx^2)^3) -  ((x^2y)^3 +(y^2z)^3 + (z^2x)^3)\cdots(1)$

we have
$a^3+b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)$
so  $((xy^2)^3 + (yz^2)^3+(zx^2)^3) = (xy^2+yz^2+zx^2)^3- 3(xy^2+yz^2)(yz^2+zx^2)(zx^2+xy^2)$
= $(xy^2+yz^2+zx^2) - 3y(xy+z^2)z(x^2+yz)x(xz+y^2)$
= $b^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(2)$
similarly
$((x^2y)^3 + (y^2z)^3+(z^2x)^3) =a^2-3xyz(x^2+yz)(y^2+xz)(z^2+xy)\cdots(3)$
from (1), (2) and (3) we get

$(x^3-y^3)(y^3-z^3)^(z^3-x^3))=b^3 - a^3 $

2015/104) Solve x^x^x^x^ to the infinity=10

x^x^x^x^ to the infinity=10

so $x^{10} = 10$

$x = \sqrt[10]{10}$

2015/103) If $xy + yz + zx =0$ then $\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy}$ is equal to

$xy + yz + zx = 0$

so $yz = -x(y+z)$
$x^2-yz = x(x+y+z)$

so $\dfrac{1}{x^2-yz} = \dfrac{1}{x(x+y+z)}$
similarly$\dfrac{1}{y^2-xz} = \dfrac{1}{y(x+y+z)}$
and
$\dfrac{1}{z^2-xy} = \dfrac{1}{z(x+y+z)}$

adding all 3 we get your expression

$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +1\dfrac{1}{z^2 - xy} = \dfrac{1}{x+y+z}(\dfrac{1}{x}+\dfrac{1}{y} + \dfrac{1}{z})$

$\dfrac{1}{x} +\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{yz+xz+xy}{xyz} = 0$

so
$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy} = 0$

2015/102) Two roots of the polynomial $x^3 + ax^2 + 15x -7 = 0$ are equal and rational. Find "a"

If 2 roots are rational then 3rd must be rational.
possible roots are -7 , -1, 1, 7
now product of 2 roots (as same) so roots shall be +1 or - 1 as 7 or -7 cannot be a double root

so 3rd root has to be 7 ( as product has to be +ve)

so $f(x) = x^3 + ax^2 + 15x -7 = 0 = f(7)$

or $343 + 49 a + 105 - 7 = 0$

or $a = 9$

check:
$x^3 - 9x^2 + 15x - 7 = (x-1)^2(x-7)$

2015/101) factor $a(b^2+c^2-a^2) + ...$(the full question below)

 $a(b^2+c^2-a^2) + b(c^2+a^2-b^2) + c(a^2+b^2-c^2) - 2abc$
Solution 

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