2017/036) Find the smallest multiple of 15 such that each digit of the multiple is either 0 or 8

Because it is multiple of 15 and 15 = 3 * 5 so it is multiple of 3 and 5. So the unit digit is zero. As 8 is not multiple of 3 there should be 3 8's for the number to be multiple of 3  and the digit cannot be the unit position. Hence the smallest number is 8880.

2017/035) a,b ,c are three distint real numbers and there are real numbers x,y such that $a^3+ax+y=0$, $b^3+bx+y=0$, $c^3+cx +y = 0$. Show that $a+b+c=0$

Let us consider the equation $f(p) = p^3 + px + y = 0$
The above equation is cubic in p and has 3 roots.
From the given condition a,b,c are three roots so the sum of roots = -  coefficient of $p^2$
hence a + b+ c = 0
Proved

2017/034) Let $p(x) = x^2 + bx +c $ where b and c are integers. if P(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4+ 4x^2 + 28x + 5$, find the value of p(1)

Because P(x) devides $x^4 + 6x^2 + 25$ and  $3x^4+ 4x^2 + 28x + 5$ hence P(x) devides the GCD,
now $GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5$
$= GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5 - 3( x^4 + 6x^2 + 25))$
$= GCD(x^4 + 6x^2 + 25, -14x^2 + 28x -70))$
$= GCD(x^4 + 6x^2 + 25, x^2 -2x+5))$ deviding 2nd expression by -14
$= GCD(x^4 + 10x^2 + 25- 4x^2, x^2-2x + 5)$
$= GCD((x^2 + 5)^2- (2x)^2, x^2-2x + 5)$
$=GCD((x^2+2x+5)(x^2-2x+5),x^2-2x  + 5) $
$=x^2 - 2x + 5$
so as p(x) is of the form $x^2+bx+ c$ so $p(x) = x^2 - 2x+5$ and hence $p(1) = 4$

2017/033) If $a+b+c+d+e+f=0$ and $a^3+b^3+c^3+d^3+e^3+f^3=0$ and no 2 variables are additive inverse of each other then show that $(a+c)(a+d)(a+e)(a+f) = (b+c)(b+d)(b+e)(b+f)$

We have $(a+c+d) = - (b+e+f)\cdots(1)$
and $a^3+c^3+d^3= -(b^3+e^3+f^3)\cdots(2)$
cube both sides of (1) using $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$  to get
$a^3+c^3+ d^3 + 3(a+c)(a+d)(d+c) = -(b^3 + e^3 + f^3 + 3(b+e)(b+f)(e+f)$
or $(a+c)(a+d)(c+d) = - (b+e)(b+f)(e+f)\cdots(3)$ using (2)
similarly we have $(a+e)(a+f)(e+f) = - (b+c)(b+d)(c+d)\cdots(4)$
multiplying (3) and (4) we get the result
 

2017/032) If $a,b,c$ are in H.P then prove that $a^3b^3 + b^3c^3 + c^3 a^3 = (9ac-6b^2)a^2c^2$

$a,b,c$ are in HP
hence $\frac{1}{a} + \frac{1}{c} =  \frac{2}{b}$
cube both sides to get  $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}(\frac{1}{a} + \frac{1}{c}) =  \frac{8}{b^3}$
or   $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}  =  \frac{8}{b^3}$
or $\frac{1}{a^3} + \frac{1}{c^3} + \frac{1}{b^3}  =  \frac{9}{b^3} -  3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}$
multiply both sides by $a^3b^3c^3$ to get
$b^3c^3 + a^3b^3 + a^3c^3 = 9a^3c^3 -  6 a^2b^2c^2 = (9aac-6b^2) a^2c^2$

2017/031) Prove that A triangle ABC is equilateral if and only if $\tan\, A + \tan\, B + \tan\, C = 3\sqrt{3}$

First let us prove the if part
if ABC is equilateral we have $\tan\, A + \tan\, B + \tan\, C = \tan\, 60^{\circ}+\tan\, 60^{\circ} + \tan\, 60^{\circ} = 3\tan\, 60^{\circ} =\sqrt{3}$
Now for the other part
we have using AM GM inequality(for all 3 positive) 
$\frac{\tan\, A + \tan\, B + \tan\, C}{3} >=\sqrt[3]{\tan\, A\,  \tan\, B\, \tan\, C}$
or $\tan\, A + \tan\, B + \tan\, C >=3\sqrt[3]{\tan\, A\,  \tan\, B\, \tan\, C}$
and these are equal if $\tan\, A = \tan\, B = \tan\, C$ or $A=B=C$ and at this  we have the sum = $3\sqrt[3]3$

2017/030)In a triangle if $\cot\,A\,\cot\,B\,\cot\,C$ are in AP then $a^2,b^2,c^2$ are in $\cdots$ progression (IIT1985-5 marks)

We have $\cot\,A\,\cot\,B\,\cot\,C$  are in AP
so
$\cot\, A + \cot\, C = 2\cot\, B$
or $\frac{\cos\, A }{\sin\, A }  +\frac{\cos\, C }{\sin\, C } = \frac{\cos\, B }{\sin\, B }$
or $\frac{\cos\, A \sin\, C + \sin\, A  \cos\, C}{\sin\, A \sin\, C } = \frac{\cos\, B }{\sin\, B} $
or $\frac{\sin ( A + C) }{\sin\, A \sin\, C }  = \frac{\cos\, B }{\sin\, B} $
or  $\frac{\sin B}{\sin\, A \sin\, C }  = \frac{\cos\, B }{\sin\, B} $ as in a triangle $\sin (A+B) = \sin\,c$
or  $\sin^2 B = 2\sin\, A \sin\, C\cos \,B $
using law of sin an cos we get
$b^2 = 2ac\frac{a^2+c^2-b^2}{ac}$
or $b^2 = a^2+c^2-b^2$
or $2b^2 = a^2 +c ^2$
hence $a^2,b^2,c^2$ are in AP

2017/029) Find the exact value for the real root of the equation $x^3+3x-2=0$

Say $f(x)=x3+3x-2=0$
$f(x)$  changes sign once so there is one positive root.
$f(-x)=-x^3-3x-2$ changes sign zero times so no -ve root
let $x=2sinh\, t$
so $ x^3+3x=2=>8\sinh\,3t+6sinh\,t=2$
or $4sinh\,3t+3sinh\,t=1$
or $sinh\,3t=1$ or $3t=sinh^{-1}1$
hence $x=2sinh(\frac{sinh^{-1}}{3})x$
this is the only real root

2017/028) Find 3 consecutive numbers that can be expressed as sum of 2 non-zero squares

We have $(2n)^2+ (2n)^2 = 8n^2$
and $(2n+1)^2 + (2n-1)^2 = 8n^2 + 2$
If we can find $8n^2 + 1  = a^2+ b^2$ for some a, b  then we are through.
we haave
$(2n - a)^2   +   (2n + a - 1)^2   =   8n^2   +   1   -   (4n - (2a^2 - 2a))$
If we can have  n   and   a   such that
$4n   -   (2a^2 - 2a)   =   0$ then we are through
or $2n   =   a^2 - a   =   a(a - 1)$
or $n = \frac{a(a-1)}{2}$

as there are infinite number of a's we find infinite nuber of solutions

a = 2, gives n= 1 2n-a = 0 so we put
a=3 , gives n= 3 we get

$6^2+6^2 = 72$
$3^2 + 8^2 = 73$
$5^2+7^2 = 74$
similarly we can find more solutions by putting any value of a.

2017/027) Find all integers n such that $(n^2-n-1)^{n+2}= 1$

There are 3 cases when
$x^y= 0$ when y = 0 or x =1 or x = -1 and y even

hence  $(n^2-n-1)^{n+2}= 1$

when
1) n+2 = 0 or n = -2
or

2) $n^2 -n -1 = 1$
or $n^2 - n -2 = 0$ or $(n-2)(n+1) = 0$ giving n = 2 or -1

or
3) $n^2 -n -1 = -1 $   and n+2  is even
in this case
$x = -1 =>n^2 -n -1 = -1$ $n^2-n=0$ so n= 0 or n =1. but n+2 need to be even so n= 0

so solution set n= 0 or 2 or -2 or -1

2017/026) if $x=\frac{4ab}{a+b}$ find the value of $\frac{x+2a}{x-2a}+ \frac{x+2b}{x-2b}$

we have
$x=\dfrac{4ab}{a+b}$
hence $\dfrac{x}{2a} = \dfrac{2b}{a+b}$
using componendo dividendo we get
 $\dfrac{x+ 2a}{x-  2a} = \dfrac{2b + (a +b)  }{2b - (a+b) } =  \dfrac{3b + a }{b- a}\cdots(1)$
similarly   $\dfrac{x+ 2b}{x-  2b} = \dfrac{2a + (a +b)  }{2a - (a+b) } =  \dfrac{3a + b}{a-b}\cdots(2)$
Adding (1) and (2) we get
 $\dfrac{x+2a}{x-2a}+ \dfrac{x+2b}{x-2b} =  \dfrac{3b + a}{b-  a}  +  \dfrac{3a + b}{a- b} $
$= \dfrac{3b + a}{b-  a}  -  \dfrac{3a + b}{b-a}$
$= \dfrac{3b + a-3a -b}{b-  a} =  \dfrac{2b-2a}{b-a} = \dfrac{2(b-a)}{b-a} = 2$

2017/025) Solve $2^y+1 = x^2$

We have
$2^y = (x^2-1)= (x+1)(x-1)$
both x+ 1 and x-1 must be power of 2. difference of them is 2.
2 powers of 2 that differ by 2 are 2 and 4

so x+ 1= 4 and x-1 = 2 or x =3 and y = 3

2017/024)If m is a root of $A(x)=x^2−nx+10=0$, and n is a root of $B(x)=x−mx+20=0$, what is the value of $(\frac{n}{m} + \frac{m}{n} )^2$

m is root of  $A(x)=x^2−nx+10=0$
so $m^2-mn+10= 0$

n is root of $B(x)=x−mx+20=0$ hence  $n^2-mn+20 = 0$

so we have $m(m-n) = - 10$ and $n(m-n) = 20$ or n = -2m

so $\frac{n}{m} = -2$ and $\frac{m}{n} = -\frac{1}{2}$ or $(\frac{n}{m} + \frac{m}{n})^2 = \frac{25}{4}$

2017/023) Find natural number n such that $n^4+33$ is a a perfect square

we need to have $n^4+33 >= (n^2+1)^2$ as next of $x^2$ is $(x+1)^2$
so $33 >= 2n^2 +1$ or $n^2 <=16$ or $n<=4$
n cannot be odd as $n^4+33$ shall be 2 mod 4 and cannot be perfect square
so n = 2 or 4
both are solutions as $2^4+33= 49 = 7^2$ and $4^4+33=289=17^2$

2017/022) Solve $(2x+1)(30x+1)(3x+1)(5x+1) = 10$

we have $(2x+1)(30x+1)(3x+1)(5x+1) = 10$
or
$(60x^2+ 32x + 1)(15x^2+ 8x + 1) = 10$

letting $15x^2 + 8x = t$
$(4t+1)(t+1) = 10$
or
$4t^2 + 5 t + 1 = 10$
or $4t^2 + 5t - 9 = 0$
or $(4t+9)(t-1) = 0 $

t =  1 or -9/4
t = 1 gives
$15x^2 + 8x-1=0$ giving $x = \dfrac{-4\pm\sqrt{31}}{15}$

or  $(15x^2 + 8x +\frac{9}4{4}) = 0$
or $(60x^2+ 32x + 9) = 0$
this gives complex solution
so solutions are  $x = \dfrac{-4\pm\sqrt{31}}{15}$

2017/021) find integer n where $\dfrac{99^n+19^n}{n!}$ is highest

Let us consider $\dfrac{99^x}{x!}$

this value increases for x upto 98 and at 99 the value is same as at 98.

now for the second part $\dfrac{19^x}{x!}$ the value increases for x upto 18 and at x = 19 is same as at x = 18 and then it decreases

for x = 19 to 98 the term $\dfrac{99^x}{x!}$ increases more rapidly as $\dfrac{19^x}{x!}$ decreases
at x = 19 the 1st term is higher by $3.8^{19}$ times and it is much higher
so we need to compare the value at x = 98 and x = 99.
the second term  is fighter at x = 98

so $\dfrac{99^x+19^x}{x!}$ is highest at x = 98

2017/020) Find n such that both $n+3$ and $n^2+3$ are perfect cubes

If $(n+3)$  and $n^2+3$ are perfect cubes then their product is a perfect cube say $x^3$

$x^3 = (n+3)(n^2+3) = n^3 + 3n^3 + 3n + 9 = (n+1)^3 + 8 = (n+1)^3 + 2^3$

$x^3 = (n+1)^3 + 2^3$ this does not have a non trivial solution

so x = 0 or n = -1 and n= -1 => n+3 is 2 which is not a cube
x = 0 => n+1 = - 2 so $n^2  + 3 =  12$ which is not a cube

so no solution

2017/019) Rational or Irrational

Let a_n be defined as follows for all natural numbers n:

a_n = 0 if the number of divisors of n (including 1 and n) is odd
a_n = 1 otherwise.

Now consider the fraction 0.a₁a₂a₃....

Is this fraction rational or irrational? Explain.

Solution:

the number of factors is odd for square number and it is even for non square numbers.

so $a_n = 0$ for no square number and = 1 for no square number

so in the decimal at each square place it is 1 and the gaps keeps on increasing.

so the digits do not recur and hence it is irrational

2017/018) ABC is a triangle and O is a point in it. Prove that $(AB + BC + AC ) > (OA + OB + OC)$

Let BO meet AC in D

Then $AB+AD > BD = OB+OD$

And $ OD+DC > OC$

Sum these for $AB+(AD+DC)+OD > OC+OB+OD$

or $AB+AC > OC+OB$

Similarly for BA+BC and CB+CA and sum to get

$2(AB+BC+CA) > 2(OA+OB+OC)$

or $AB+BC+CA > OA+OB+OC$

2017/017) Find positive n such that $lim_{x->3} \frac{x^n-3^n}{x-3} = 108$

we have from definition LHS = $\frac{dy}{dx}$ at x = 3 where $y=x^n$
so differentiating we get $(n)3^{n-1} = 108 = 4 * 3^3$ or n = 4

2017/016) Find a,b satisfying $4^a + 4a^2 + 4 = b^2$

The LHS is even hence to put a limit on a we must have
$(2^a+2)^2 <= 4^a + 4a^2 + 4$
or $4^a + 4 * 2^a + 4 <= 4^a + 4a^2 + 4$ or $2^a <= a^2$ or $ a <= 4$
putting a = 0 to 4 one by one we get $a=2,b=6$ and $a=4,b= 18$ are 2 solution

2017/015) Show that there are infinitely many n such that $6n+1$ and $6n-1$ are both are composites

we have for n > 2 $n^3-1= (n-1)(n^2+n+1)$ is composite and $n^3+1=(n+1)(n^2-n+1)$ is composite for n ,
so if we  choose $n= 36m^3$ we get $6n-1= (6m)^3-1$ and $6n+1= (6m)^3 + 1$ composites

2017/014) if $x,y,z$ are in HP then show that $\log(x+z) + \log(x+z-2y) + 2 \log(x-z)$(IIT-1978)

let us eliminate y from LHS as RHS does not contain y
We have $x,y,z$ are in HP
hence $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$
or $y(z+x) = 2xz$
or $y=\frac{2xz}{x+z}$
hence $x+z-2y = x+ z - \frac{4xz}{x+z}$
or $(x+z-2y)(x+z) = (x+z)^2 - 4xz = x^2+z^2+ 2xz - 4xz$
$ = x^2+z^2 - 2xz = (x-z)^2$
taking log on both sides we get $\log(x+z) + \log(x+z-2y) + 2 \log(x-z)$

2017/013) Prove that $\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$

we know $\sin \, nt = \frac{1}{2i}(e^{int} - e^{-int})=\frac{1}{2i}e^{-int}(e^{2int}-1) $
lettiing $t=\frac{\pi}{m}$ we have taking the product from n = to m-1 we get
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}}\prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1) \cdots(1) $

Now $ e^{-i\sum_{n=1}^{m-1}\frac{n\pi}{m}} = e^\frac{-(m-1)(m)\pi\,i }{2m}= e^\frac{-(m-1)\pi\,i }{2}=  (e^\frac{-\pi\,i}{2})^{(m-1)}= (-i)^{m-1}\cdots(2)$

from (1) and (2)
$\prod_{n=1}^{m-1} sin \frac{n \pi}{m}= (\frac{1}{2i})^{m-1} (-i)^{m-1}= (\frac{-1}{2})^{m-1} \prod_{n=1}^{m-1} (e^{2\frac{in\pi}{m}}-1)$
$= (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- e^{2\frac{in\pi}{m}}) = (\frac{1}{2})^{m-1} \prod_{n=1}^{m-1} (1- w^n) \dots (3)$ where w is $m^{th}$ root of 1

now because w is $n^{th}$ root of 1 we have

$x^n-1 = \prod_{n=0}^{m-1} (x- w^n) = (x-1)  \prod_{n=1}^{m-1} (x- w^n)\cdots(4)$
further $x^n-1 = (x-1) \sum_{n=0}^{m-1} x^n\cdots(5)$
so we have from (4) and (5)
$\prod_{n=1}^{m-1} (x- w^n) = \sum_{n=0}^{m-1} x^n$
putting x = 1 we get
$\prod_{n=1}^{m-1} (1- w^n) = \sum_{n=0}^{m-1} 1^n= m $

from (3) and above we get
$\prod_{n=1}^{m-1} \sin \frac{n \pi}{m} = \frac{m}{2^{m-1}}$

2017/012) Can $5^{64}-3^{64}$ be expressed as sum of 2 squares

we have $5^{64}-3^{64}= (5^{32} + 3^{32})(5^{32} - 3^{32})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{16} - 3^{16})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{4} - 3^{4})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{2} + 3^{2})(5^{2} - 3^{2})$
$= (5^{32} + 3^{32})(5^{16} + 3^{16})(5^{8} + 3^{8})(5^{4} + 3^{4})(5^{2} + 3^{2}) * 16$
the above is product of sum of squares( knowing that $16= 4^2 +0^2$ the above can be expressed as sum of squares

2017/011) If the sum of first n terms of an A.P. is $cn^2$ find the sum of squares of these n terms.

the sum of $1^{st}$ n terms is $cn^2$
so $k^{th}$ term is $ck^2-c(k-1)^2= c*(2k-1)$
so sum of n terms = $\sum_{k=1}^{n} (c^2(2k-1)^2) = c^2 \sum_{k=1}^{n} (4k^2-4k + 1) = c^2(4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2} +n)$
= $\frac{nc^2(4n^2-1)}{3}$

2017/010) Let $a,\,b,\,c,\,d$ be real with condition that $a+b\sqrt{2}+c\sqrt{3}+2d >= \sqrt{10(a^2+b^2+c^2+d^2)}$

show that $a^2+d^2=b^2+c^2$

Proof:

by Cauchy-Schawartz in equality we have
$(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $\sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)$
from given condition and above we have
$\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
when
$\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)$
So $a = k, b^2= 2k^2,c^2=3k^2,d=2k$ and hence $a^2+d^2=b^2+c^2$

2017/009}Prove that $\sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}\lt 2016$

we have $n^{th}$ term = $\frac{\sqrt{n\cdot (n+1)}}{2n+1}$
$= \frac{\sqrt{n^2+n}}{2n+1}$
$= \frac{\sqrt{n^2+n+\frac{1}{4}-\frac{1}{4}}}{2n+1}$
$=  \frac{\sqrt{(n+\frac{1}{2})^2-\frac{1}{4}}}{2n+1}$
$ < \frac{n+\frac{1}{2}}{2n+1}$
$ < \frac{1}{2}$
each term is $ < \frac{1}{2}$ and there are 4032 terms so sum is less than 2016

2017/008) Let P be a polynomial such that $P(x)=a_0+a_1x+?+a_nx^n$where $a_0,a_1,\cdots$? are non-negative integer. Given that P(1)=4 and P(5)=152 find P(6)

P(x) is a cubic polynomial as P(5) is 152 between $125(5^3)$ and $625(5^4)$
now coefficient of $x^3$ is 1
so $P(x) = x^3 + ax^2 + bx + c$
taking mod 5 we get c =2 so $P(X) = x^3 + ax^2+ bx + 2$ and so $a . 5^2 + 5b = 152-125-2 = 27$ giving a = 1, b= 2
sp $P(x) = x^3 +x ^2 + 2$ and it satisfies P(1) = 4
so P(6) = 254

2017/007) if $\alpha,\beta$ are the roots of $x^2-5x+1$ show that $\alpha^n+\beta^n$ is an integer and not divisible by 4 for any integer n

we have for n = 1 $\alpha+\beta = 5 $ which is integer not divsible by 4
and $\alpha\beta=1$
hence $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 5^2 - 2 = 23$ integer and not divisible by 4
hence it is true for n =1 and 2
let it be true for n = 1 to k for k >=2
now
$( \alpha^n+\beta^n)(\alpha+\beta) = \alpha^{n+1}+\beta^{n+1} + (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
or $\alpha^{n+1}+\beta^{n+1} = 5 (\alpha^n+\beta^n) - (\alpha^{n-1} + \beta^{n-1}$
or $\alpha^{n+1}+\beta^{n+1} =  4(\alpha^n+\beta^n) +  (\alpha^n+\beta^n)- (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
Now let n be smallest integer so that $\alpha^{n}+\beta^{n}$ is divisible by 4
so $\alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2}$ is divsible by 4
so $\alpha^{n-3}+\beta^{n-3} = alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2} - 4alpha^{n-2}+\beta^{n-2}$
is divisible by 4 which is contadiction.
hence proved

2017/006) if $x= a + b $ $y = aw+bw^2$ and $z= aw^2+ bw$ then show that $x^3+y^3 + z^3 = 3(a^3+b^3)$ and $x^2+y^2 + z^2 = 6ab$

We have $x+y+z = a(1+w+w^2) + b(1+w+w^2) = 0$
so x^3 + y^3+ z^3 = 3xyz = 3(a+b)(aw+bw^2)(aw^2+bw) = 3(a+b)(a^2 w^3 + abw^2 + abw^4 + b^2w^3)$
= 3(a+b)(a^2 + ab(w^2+ w^4) + b^2) = 3(a+b)(a^2 + ab(w^2+w) + b^2)$
$ = 3(a+b)(a^2 - ab + b^) = 3(a^3+b^3)$
Further
$x^2 = a^2 + b^2 + 2ab$
$y^2 = a^2w^2 + b^2w + 2ab$
$z^2 = a^2w + b^2w^2 + 2ab$
hence $z^2+y^2+z ^2 = a^2(1+w+w^2) + b^2(1+w+w^2) + 6ab = 6ab$

2017/005) Solve $|z+1| = z+ 2 + 2|$

because LHS is real so z + 2i is real and let z + 2i = x or z = x - 2i
so |x+1 - 2i |= x +2|
or $(x+1)^2 + 4 = (x+2)^2$
or $x^2 + 2x + 5 = x^2 + 4x + 4$ or $x = \frac{1}{2}$
$z = \frac{1}{2} - 2i$    

2017/004) If $x^2+px+1$ ia a factor of $ax^3 + bx +c $ then

a) $a^2+c^2= -ab$, b) $a^2-c^2 = - ab $, c) $a^2-c^2 = ab$,  d) None of these

Solution
we have other factor linear and hence $dx + e$
so $ax^3+ bx + c = (x^2+px+1)(dx + e)$
comparing coefficient of $x^2$ we have d = a and constant term gives c =e
so  $ax^3+ bx + c = (x^2+px+1)(ax + c) =  ax^3 + x^2(c + ap) + x(a + cp) + x$
comparing coefficients of $x^2$ on both sided $c+ap=0\cdots(1)$
comparing coefficient of x both sides $ a+cp = b$ or  $a^2 + acp = ab$ or $a^2-c^2 = ab$(using from (1)) Hence C

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 $a > \frac{9}{2}$