2017/003) Show that for a,b,c all different $\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} != 0$

if  $\sqrt[3]{a-b} + \sqrt[3]{b-c } + \sqrt[3]{c-a} = 0$
then using $(x + y + z = 0) =>x^3 + y^3 + z^3 = 3xyz$
we get $0 = 3\sqrt[3]{(a-b)(b-c)(c-a)}$ giving a = b or b= c or c=a.
so if all are different result cannot be zero.

2017/002) Given a cubic eqaution $x^3 + px^2+ qx+r =0$ has 2 roots

 that are equal in value and opposite in sign show that $pq=r$

Proof:

Let the roots be $a,-a,b$
so we have $x^3+px^2+qx + r = (x-a)(x + a)(x-b) = (x^2 - a^2)(x-b) = x^3 - b x^2 -a^2 x + ba^2$
giving $p = -b$, $q=-a^2$, $r = ba^2$
hence $pq = a^2b = r$

2017/001) Find all values of the parameter a such that the real roots $\alpha,\beta$ of the equation $2x^2+6x+a=0$ satisfy the ineqauilty $|\frac{\alpha}{\beta} + \frac{\beta}{\alpha} |< 2$

as  for real t $|t+\frac{1}{t}| >=2$
so one root is positive and another -ve
$2x^2+6x + a = 0$
$=>4x^2 + 12x + 2a = 0$
$=>(2x+3)^2 + 2a - 9 = 0$
$=>(2x+3)^2 =  9 - 2a$
$x = -\frac{3}{2} \pm \frac{\sqrt{9-2a}}{2}$
for one root to be positive and another -ve $\frac{\sqrt{9-2a}}{2} > \frac{3}{2}$ or $ a < 0$
if we were interested in complex roots then $a > \frac{9}{2}$