\(\sqrt { 16{ x }^{ 2 }+9{ y }^{ 2 }-24xy+24x-18y+9 } \)
= \(\sqrt { { \left( 4x \right) }^{ 2 }+{ \left( -3y \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+2{ \left( 4x \right) }\left( -3y \right) +2\left( -3y \right) \left( 3 \right) +2\left( 4x \right) \left( 3 \right) } \)
= \(\sqrt { { \left( 4x+-3y+3 \right) }^{ 2 } } \) = |4x - 3y + 3|

Let f(x) = 2x³ - 5x² + 5x - 3 and g(x) = x³ + x² - x + 2

-7(x² - x + 1) ≠ 0, note that -7 is not a divisor of g(x)
Now dividing, g(x) = x³ + x² - x + 2 by the new remainder x² - x + 1 (leaving the constant factor), we get

Here, we get zero remainder
Therefore, GCD (2x³ - 5x² + 5x - 3, x³ + x² - x + 2) = x² - x + 1.

Let \(\frac {1}{x}\) = p, \(\frac {1}{y}\) = q, \(\frac {1}{z}\) = r
The given equations are written as
\(\frac { p }{ 2 } +\frac { q }{ 4 } -\frac { r }{ 3 } =\frac { 1 }{ 4 } \)
p = \(\frac {q}{3}\)
p - \(\frac {q}{5}\) + 4r = 2\(\frac {2}{15}\) = \(\frac {32}{15}\)
By simplifying we get,
6p + 3q - 4r = 3...(1)
3p = q          ...(2)
15p - 3q + 60r = 32                  ..(3)
Substituting (2) in (1) and (3) we get,
15p - 4r = 3             ....(4)
6p + 60r = 32 reduces to 3p + 30 r = 16.... (5)
Solving (4) and (5),

Substituting r = \(\frac {1}{2}\) in (4) we get, 15p - 2 = 3 gives, p = \(\frac {1}{3}\)
From (2), q = 3p  we get q = 1
Therefore, x  = \(\frac {1}{p}\) = 3, y = \(\frac {1}{q}\) = 1, z = \(\frac {1}{r}\) = 2. That is, x = 3, y = 1, z = 2.

Let the present age of father be x years and the present age of son be y years
Given, x = 6y … (1)
x + 6 = 4(y + 6) … (2)
Substituting (1) in (2), 6y + 6 = 4(y + 6)
6y + 6 = 4y + 24 gives, y = 9
Therefore, son’s age = 9 years and father’s age = 54 years.

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

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

\(2{ x }^{ 2 }-2\sqrt { 6 } x+3=2{ x }^{ 2 }-\sqrt { 6 } x-\sqrt { 6 } x+3\) (by splitting the middle term)
= \(\sqrt { 2 } x\left( \sqrt { 2 } x-\sqrt { 3 } \right) -\sqrt { 3 } \left( \sqrt { 2 } x-\sqrt { 3 } \right) =\sqrt { 3 } \left( \sqrt { 2 } x-\sqrt { 3 } \right) =\left( \sqrt { 2 } x-\sqrt { 3 } \right) \)
Now, equating the factors to zero we get,
\(\left( \sqrt { 2 } x-\sqrt { 3 } \right) \)\(\left( \sqrt { 2 } x-\sqrt { 3 } \right) \) = 0
\( \sqrt { 2 } x-\sqrt { 3 } \) = 0 
 \(\sqrt { 2 } x=\sqrt { 3 } \)
Therefore the solution is x = \(\frac { \sqrt { 3 } }{ \sqrt { 2 } } \).