2015/114) Solve in real $a^3 + b^3 = 8 – 6ab$

$a^3 + b^3 - 8 + 6ab = 0$

or

$a^3 + b^3 + (-2)^3 -3(-2)(a)(b) = 0$

=> $a + b – 2 = 0$ or $a=b= - 2$

using the fact

$x^3 + y^3 + z^3 – 3xyz = \dfrac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

2015/113) There is a number n between to successive squares.this is k larger than the smaller number and l smaller than the larger number . Prove that n-kl is a perfect square.

Let the number n be between $a^2$ and $(a+1)^2$
As per given condition
$n- a^2 = k\cdots(1)$
$(a+1)^2 –n = l\cdots(2)$
Adding (1) and (2)
$k + l = 2a + 1$
or $l = 2a + 1 - k$
now
$n- kl = (a^2+k) – k(2a+1-k)
= a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2$
Hence proved

2015/112) show that $\cos2A + \cos6A + \cos8A = \dfrac{\sqrt{13)}-1}{4}$ where $A = \dfrac{pi}{13}$

Let
$x = \cos2A + \cos6A + \cos8A \cdots(1)$
by seeing that $\sqrt{13}$ on right
square both sides of (1) to get
$x^2 = \cos^2 2A + \cos^2 6A + \cos^2 8A + 2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A$
multiply by 2 to get
$2 x^2 = 2 \cos^2 2A + 2 \cos^2 6A + 2 \cos^2 8A$
        + $2(2 \cos 2A \cos 6A + 2 \cos 2A \cos 8A + 2 \cos 6A \cos 8A)$
= $\cos 4A + 1 + \cos 12 A + 1 + \cos 16 A + 1$ 
                    + $2( \cos 8A +\cos 4 A + \cos 10 A + \cos 6A + \cos 14 A + \cos 2 A)$
= $3 + \cos 4A + \cos 12 A + \cos 16 A$
                      + 2 $( \cos 8A + \cos 6A + \cos 2A + \cos 4A + \cos 10 A + \cos 14A)$

Now $\cos 16 A = \cos 10 A$ as $26 A = 2\pi$
$\cos 14 A = \cos 12 A$ as $26 A = 2\pi$
So we continue
= $3 + \cos 4A + \cos 12 A + \cos 10 A + 2(x+ \cos 4A + \cos 10 A + \cos 12 A)$
= $3 + 2x + 3 (\cos 4A + \cos 12 A+ \cos 10 A)$
Now $\cos 2A + \cos 4A + \cos 6A + \cos 8A + \cos 10 A + \cos 12A = \dfrac{-1}{2}$

So $\cos 4A + \cos 12 A+ \cos 10 A = (\dfrac{- 1}{2}-x)$

So $2x^2 = 3 + 2x + 3(\dfrac{-1}{2}- x)$
Or $4x^2 = 6 + 4x -3 – 6x$
Or $4x^2 + 2x -3 = 0$

This has one positive solution $\dfrac{\sqrt{13)}-1}{4}$ and one negative solution

As $\cos2A + \cos6A + \cos8A = \cos 2A + \cos 6A – \cos 5A$ and $\cos6 A > 0$ and $\cos 2A > \cos 5A$ so this is > 0
So this is $\dfrac{\sqrt{13)}-1}{4}$


I have solved at http://in.answers.yahoo.com/question/index;_ylt=ApwbfmdwbadqGq5DHY.KZsiRHQx.;_ylv=3?qid=20120607073236AAFCjFo

2105/111) Given 4 positive integers a,b,c and d such that $a^5=b^4, c^3=d^2$ and $c−a=19$ what is $d−b$

as $c^3 = d^2$ so c will be a square let $c = x^2$

as $a^5 = b^4$ so $a = y^4$

now
$c-a = 19$
=> $x^2 - y^4 = 19$
=> $(x-y^2)(x+y^2) = 19$
hence $x - y^2 =1$ and $x+y^2 = 19$ as 19 is a prime
so $x = 10$ and $y = 3$

so $a = y^4$ or $a^5 = y^20 = b^ 4$ or $b= y^5 = 243$

$c= x^2 => c= 100$ and hence $d^2 = 10^6$ and so $d = 1000$
$d-b = 1000 - 243 = 757$

2015/110) If 'a' and 'b' are the roots of $x^2-3x+1=0$ then

find the value of $\frac{a^{2014} + b^{2014} + a^{2016} + b^{2016}}{ a^{2015} + b^{2015}}$

Solution

a is root of $x^2 – 3x + 1=0$

so $a^2 - 3a + 1 = 0$

or

$a^2 + 1 = 3a$

so

$\dfrac{a^{2014} + a^{2016}}{a^{2015}} = \dfrac{1+a^2}{a} = 3 \cdots(1)$

Similarly

$\dfrac{b^{2014} + b^{2016}}{b^{2015}} = \dfrac{1+b^2}{b} = 3 \cdots(2)$

using

if $\dfrac{x}{y} = \dfrac{z}{w}$ then both are $\dfrac{x+z}{y+w}$

we get

$\dfrac{a^{2014} + b^{2014} + a ^{2016} + b^{2016}}{a^{2015} + b^{2015}} =  3$

2015/109) A triangle with sides 10, 24, and 26 is inscribed in a circle. What is the radius of the circle?

This is a right angled triangle because $10^2 + 24^2 = 26^2$ ( as u can check)
Therefore the hypotneuse forms the diameter and therefore the radius is $\frac{26}{2} =13$

2015/108) If $a + b = 1$ and $a^2 + b^2 = 2$ what is the value of $a^3 + b^3$ ?

we have

$a+b =1 \cdots(1)$
$a^2+b^2 = 2 \cdots(2)$
from 1st we get
$(a+b)^2 = a^2 +b^2 + 2ab = 2 + 2ab = 1$
or $ab = \dfrac{- 1}{2}$

hence
$a^3+b^3 = (a+b)^3 - 3ab(a+b) = 1 - 3 * \dfrac{-1}{2} * 1 = \dfrac{5}{2}$