Squaring both sides(\(\sqrt { y+1 } +\sqrt { 2y-5 } \) )² = 3²
y+1+2y-5+2(\(\sqrt { y+1 } +\sqrt { 2y-5 } \)) = 9
3y-4-9 = -2\(\sqrt { y+1 } +\sqrt { 2y-5 } \)
Again squaring both sides
(3y-13)²(-2\(\sqrt { y+1 } +\sqrt { 2y-5 } \))²
9y²-78y+169 = 4(y+1)(2y-5)
9y²-78y+169 = 4(2y²+2y-5y-5)
9y²-78y+169 = 8y²+8y-20y-20
9y²-78y+169-8y²+12y+20 = 0
y²-66y+189 = 0
y²-63-3y+189 = 0
y(y-63)-3(y-63) = 0
(y-63)(y-3) = 0
y = 63, 3

|17x² - 18x + 19|

Let Arul's speed of working be x
Let Mohan's speed of working be y
Let Ram's speed of working be z
given that they are working together.
Let 'w' be the quantum of work.
Also given that Mohan takes twice the time as Arul for finishing the work.
\(\therefore \frac { w }{ y } =2\times \frac { w }{ x } \therefore x=2y\)
\(\therefore y=\frac { x }{ 2 } \)   (2)
Also Ram takes 3 times the time as Arul for finishing the work.
\(\therefore \frac { w }{ z } =3\times \frac { w }{ x } \)
\(\therefore x=3z\quad \therefore z=\frac { x }{ 3 } \)
Substitute (2) and (3) in (1),
\(x+\frac { x }{ 2 } +\frac { x }{ 3 } =\frac { w }{ 6 } \)
∴ 6x + 3x + 2x = w
11x = w
\(x=\frac { w }{ 11 } ,y=\frac { w }{ 22 } ,z=\frac { w }{ 33 } \)
Working alone time taken as
\(Arul:\frac { w }{ x } =\frac { w }{ w/11 } =11hrs.\)
\(Mohan:\frac { w }{ y } =\frac { w }{ w/22 } =22hrs.\)
\(Ram:\frac { w }{ z } =\frac { w }{ w/33 } =33hrs\)

\(=\frac { (q+r)+p }{ (q+r)-p } \times \frac { (q+r)+p }{ (q+r)+p } \times \frac { 2qr+{ q }^{ 2 }+{ r }^{ 2 }-{ p }^{ 2 } }{ 2qr } \)

\(=\frac { { (q+r+p) }^{ 2 } }{ 2qr } =\frac { 1 }{ 2qr } \)

\(\frac { { x }^{ 3a }-8 }{ { x }^{ 2a }+2{ x }^{ a }+4 } \)
\(=\frac { { ({ x }^{ a }) }^{ 3 }-8 }{ { ({ x }^{ a }) }^{ 2 }+{ 2x }^{ a }+4 } =\frac { { ({ x }^{ a }) }^{ 3 }-{ 2 }^{ 3 } }{ { x }^{ 2a }+{ 2x }^{ a }+4 } \)
\(=\frac { ({ x }^{ a }-2)\left( { x }^{ 2a }+2{ x }^{ a }+4 \right) }{ { x }^{ 2a }+{ 2x }^{ a }+4 } \)
\(=\frac { ({ x }^{ a }-2)\left( { x }^{ 2a }+2{ x }^{ a }+4 \right) }{ { x }^{ 2a }+{ 2x }^{ a }+4 } ({ x }^{ a }-2)\)

Let f(x) = 2x⁴ + 13x³ + 27x² + 23x + 7,
g(x) = x³ + 3x² + 3x + 1,
h(x) = x² + 2x + 1
which is the least degree polynomial.
Now Dividing f (x) by h(x)

Since the Remainder is zero, h (x) is the GCD of f (x) and h (x)
Now, dividing g (x) by h (x)

Remainder is zero, h (x) is the GCD of g (x) and h (x)
h (x) divides both f(x) and g (x) completely
GCD = x² + 2x + 1

xy (k² + 1) + k(x², y²)
Factorizing xy k² + xy + kx² + ky²
= xyk² + ky² + kx² + xy
= ky ( kx + y) + x ( kx + y)
= (kx + y) ( ky + x)
and xy (k² - 1) + k(x² - y²)
= xyk² - xy + kx² - ky²
= xyk² + kx² - xy - ky²
= kx (ky + x) - y (x + ky)
= (kx + y) (kx - y)
LCM = (ky + x) (kx + y) (kx - y)
= (ky - x) (k² x² - y²)

Let the 
100's digit be 'x'
10's digit be 'y'
Unit's digit be 'z'
Given 100y+ 10x + z - 54 = 3(100x + 10y + z)
Substituting 290x - 70y + 22 = -54 ( /2)
145x - 35y + z = -27          .........(1)
l00x + 10y + z + 198 = 100z + 10y + x
Substituting 99x - 992 = - 198 (99)
x - z = - 2              ...........(2)
y = x + 2 (y - z)
x + y - 2z = 0          ..........(3)
Consider (1) and (3)
145x - 35y + z = -27        .......(1)
35x + 35y - 70z = 0          .......(4)
180x - 69z = -27            .......(5)
Consider (5) and (2)

\(x=\frac{111}{111}=1\)
Substituting x = 1 in         .........(2)
1 - z = -z
z = 1 + 2 = 3
Substituting x - 1, z = 3 in (3)
1 + y - 6 = 0
y = 5
solution: x = 1, y = 5, z = 3
The number is 153.

Let the number of students in sections A, B and C be 'x', 'y' and 'z' respectively
Given x + y + z = 150           ........(1)
x - 6 = z + 6
x - z = 12            .....(2)
4z = x + y
x + y - 4z = 0         ..........(3)
consider (1) and (3)

\(z=\frac{150}{5}=30\)
substituting in (2)
x - 30 = 12
x = 30 + 12
x = 42
substituting x = 42, z = 30 in (1)
42 + y + 30 = 150
y = 150 - 72
y = 78
The number of students in sections A, B and C are 42, 78, 30 respectively.

\(\frac{1}{3}(x+y-5)=y-z\)
x + y - 5 = 3y - 3z
x - 2y + 3z = 5          ..............(1)
y - z = 2x - 11
2x - y + z = 11     ........(2)
2x - 11 = 9 - (x + 2z)
2x + x + 2z = 9 + 11
3x + 2z  = 20             ......(3)
Consider (1) and (2)

z = 3/3 = 1
Substituting z = 1 in (3)
3x + 7 (1) = 20
3x = 20 - 2 = 18
\(x=\frac{18}{3}=6\)
Substituting x = 6,z = 1 is                   ...........(1)
6 - 2y + 3(1) = 5
6 + 3 - 5 = 2y
4 = 2y
y = 2y
\(y=\frac{4}{2}=2\)
Solution: x = 6,y = 2, z = 1.