2021/012) If a and b are roots of $x^4 + x^3 - 1$ then prove that ab is a root of $x^6 + x^4 + x^3 - x^2 - 1$

 let the other 2 roots be c and d

then we get comparing coefficients (this is given by viete;s relation also)

a+b+c+d = -1

ab+ac+ad+bc+bd+cd = 0

abc+abd + acd + bdc = 0

abcd = -1

putting a+ b= s c+d = t, ab = p and cd = q we get

s+ t = -1  ...1

p + q + st = 0....2 

pt + qs = 0...3 

and pq = -1...4

$q = - \frac{1}{p}$ (from 4) and t = -1 -s(from 1) we get

$p - \frac{1}{p} - s^2 -s = 0 \cdots(5)$ 

and $p(-1-s) - \frac{s}{p} = 0 => p^(1-s) - s  = 0$

$=> s =\frac{p^2}{1+p^2} \cdots(6)$

from 6 put s in 5 to get

$p - \frac{1}{p} -\frac{p^4}{(p^2+1)^2} +\frac{p^2}{(1+p^2} = 0$

or $p^2(p^2+1)^2 -(p^2+1)^2 -p^5 +p^3(p^2+1) = 0$

or $p^2(p^4+2p^2+1) - (p^4 + 2p^2 + 1)  -p^5 + p^5 + p^3 = 0$

or $p^6 +p^4 +p^3 -p^2 -1 = 0$

so p is a root of $x^6 +x^4 + x^3 -x^2 -1$ 

so ab (which is p)  a root of $x^6 +x^4 + x^3 -x^2 -1$

2021/011) Given positive a,b,c show that $a^2(b+c) + b^2(a+c) + c^2(a+b) >= 6abc$

 We have 

LHS $= abc(\frac{a}{c} + \frac{a}{b}) +  abc(\frac{b}{c} + \frac{b}{a})+ abc(\frac{c}{b} + \frac{c}{a})$

$= abc(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + \frac{b}{a} + \frac{c}{b} + \frac{c}{a})$

$= abc(\frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{b} + \frac{b}{a})$

as $\frac{a}{c} + \frac{c}{a} \ge 2 $ 

$\frac{b}{c} + \frac{c}{b} \ge 2 $

$\frac{a}{b} + \frac{b}{a} \ge 2 $

The above 3 relations follow from am gm inequality

putting the value we get

LHS $>= abc(2+2+2)$ or LHS $>= 6abc$ 

2021/010) Simplify $\cot 70^\circ +4\cos70^\circ=$

 $\cot 70^\circ +4\cos70^\circ=$

$=\frac{\cos 70^\circ}{\sin 70^\circ} +4\cos70^\circ=$

$=\frac{\cos 70^\circ + 4\sin 70^\circ \cos 70^\circ}{\sin 70^\circ }$

$=\frac{\cos 70^\circ +2(2\sin 70^\circ \cos 70^\circ)}{\sin 70^\circ }$

$=\frac{\cos 70^\circ +2\sin 140^\circ}{\sin 70^\circ }$

$=\frac{\sin 20^\circ +2\sin 40^\circ}{\sin 70^\circ }$

$=\frac{(\sin 20^\circ +\sin 40^\circ) + \sin 40^\circ}{\sin 70^\circ }$

$=\frac{(2 \sin 30^\circ \cos 10^\circ) + \sin 40^\circ}{\sin 70^\circ }$ using sin A + sin B formula

$=\frac{\cos 10^\circ  + \sin 40^\circ}{\sin 70^\circ }$ 

$=\frac{\sin 80^\circ + \sin 40^\circ}{\sin 70^\circ }$ 

$=\frac{2  * \sin 60^\circ  \cos 20^\circ}{\sin 70^\circ }$ 

$=\frac{\sqrt{3} \sin 70^\circ}{\sin 70^\circ }$

$=\sqrt{3}$

2021/009) Solve for integers $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

Because of symmetry we have if $a,b,c$ is a solution then any permutation is a solution

Without loss of generality we choose $a \le b \le c$

So $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})(1+\frac{1}{a})(1+\frac{1}{a})$

Or  $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) \le (1+\frac{1}{a})^3$

So from given condition $(1+\frac{1}{a})^3 \ge 2$ or $(\frac{a+1}{a})^2 \ge 2$

The LHS is decreasing function of a and we have to see ratio of 2successsive cubes >2 as equal to 2 is not possibe

we have $\frac{64}{27} > 2$ and  $\frac{125}{64} <2$ giving $a < 4$

So we can have upto 3 values(as for some value we may not have a solution) for a . that is 1,2,3

putting a = 1 we get  $2(1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c})= 1$

as for b and c $\ge 1$ LHS is less than 1 this does not have a solution

putting a = 2 we get  $\frac{3}{2} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $(1+\frac{1}{b})(1+\frac{1}{c}) = \frac{4}{3}$

or $3(b+1)(c+1) = 4bc$

or $3bc + 3b + 3c + 3 = 4bc$

or $bc -3b -3c -3 = 0$

or $b(c-3) -3 (c-3) = 12 $

or $(b-3)(c-3) = 12 = 1 * 12 = 2 * 6 = 3 * 4$

by taking one set a a time(as we have $b \le c$  we get (b,c) = (4,15) one solution (5,9) another solution and (6,7) 3rd solution

putting a = 2 we get  $\frac{4}{3} (1+\frac{1}{b})(1+\frac{1}{c}) = 2$

or  $4 (1+\frac{1}{b})(1+\frac{1}{c}) = 2 * 3$

or $2 (1+\frac{1}{b})(1+\frac{1}{c}) = 3$

or $2(b+1)(c+1) = 3bc$

or $2bc + 2b + 2c + 2 = 3bc$

or $(bc-2b -2c = 2$

or $(b-2)(c-2) = 6 = 1 * 6 = 2 * 3$

giving 2 sets of solutions (b,c) as (3,8) or (4,5)

So we have full solution in (a,b,c) as (2,4,15), (2,5,9),(2,6,7), (3,3,8), (3,4,5) or any permutation of any of the solutions 

2021/008) Given $\frac{x}{y+z} + \frac{y}{z+ x} + \frac{z}{x+y} = 1 $ Find $\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} $

Because 1st term of the LHS in the expected result has $x^2$ we need to multiply 1st term of the given expression by y , 2nd term by y and 3rd tert by z so we can multiply by x+y + z to get

$\frac{x*(x+y+z)}{y+z} + \frac{y*(x+y+z)}{z+ x} + \frac{z*(x+y+z)}{x+y} = 1 $

Or

$\frac{x^2}{y+z} + x  + \frac{y^2}{z+ x} + y + \frac{z^2}{x+y} + z = x+ y + z$

Or 

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} + x + y + z = x+ y + z$

Or

$\frac{x^2}{y+z} + \frac{y^2}{z+ x} + \frac{z^2}{x+y} = 0$

2021/007) Three real numbers are given, Fractional part of product of every 2 of them is $\frac{1}{2}$. Prove that these numbers are irrational.

 Let the three numbers be a,b,c as product of every 2 numbers has fractional part $\frac{1}{2}$ so each of the product is an integer plus $\frac{1}{2}$ 

So there exists integers p,q r such that $ab = \frac{2p+1}{2}\cdots(1)$

 $bc = \frac{2q+1}{2}\cdots(2)$ 

 $ca = \frac{2r+1}{2}\cdots(3)$ 

 multiplying all 3 we get $(abc)^2 = \frac{(2p+1)(2q+1)(2r+1)}{8}$ 

 or $(abc) = \frac{\sqrt{(2p+1)(2q+1)(2r+1)}}{2\sqrt{2}}$ 

 Now numerator is square root of odd number and denominator is product of 2 and$\sqrt{2}$ so the number is irrational and as product of all 3 is irrational and product of 2 number is rational so 3rd number is irrational.

 this is true for each pair and hence all 3 numbers are irrational. Hence proved