\({ \left( 729 \right) }^{ \frac { -1 }{ 6 } }={ \left( { 3 }^{ 6 } \right) }^{ \frac { -4 }{ 6 } }={ 3 }^{ -4 }\)
=\(\frac{1}{3}\)

3√45 - √125 + √200 - √50
= 9√5-5√5 + 10√2 - 5√2
= 4√5 + 5√2

If possible let √2 be rational and let its simplest form be \(\frac{a}{b}\) where a and b are integers having no common factor other than 1 and b≠0
Now, √2 = \(\frac{a}{b}\)⇒2 = \(\frac{a^{2}}{b^{2}}\) [on squaring both sides]
⇒ 2b² = a² ...(1)
⇒ 2 divides a² 
⇒ 2 divides a
Let a = 2c for some integer c.
Putting a = 2c in (1), we get
2b² = 4c² ⇒ b² = 2c²
⇒ 2 divides b²
⇒ 2 divides b
Thus 2 is a common factor of a and b
But this contradicts the fact that a and b have no common factor other than 1 the contradiction arise by assuming that √2 is rational.
Hence √2 is irrational.

\(\frac { 1 }{ 6 } (\sqrt { 3 } -6)\)

2

\(\frac { 97393 }{ 990 } \)

\(\frac { 4 }{ 2+\sqrt { 3 } +\sqrt { 7 } } \)
\(=\frac { 4 }{ 2+\sqrt { 3 } +\sqrt { 7 } } \times \frac { (2+\sqrt { 3 } -\sqrt { 7 } ) }{ (2+\sqrt { 3 } -\sqrt { 7 } ) } \)
\(=\frac { 8+4\sqrt { 3 } -4\sqrt { 7 } }{ { (2+\sqrt { 3 } ) }^{ 2 }-{ (\sqrt { 7 } ) }^{ 2 } } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { 8+4\sqrt { 3 } -4\sqrt { 7 } }{ 4+3+4\sqrt { 3 } -7 } \ \left[ \because \ { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=\frac { 8+4\sqrt { 3 } -4\sqrt { 7 } }{ 7+4\sqrt { 3 } -7 } =\frac { 8+4\sqrt { 3 } -4\sqrt { 7 } }{ 4\sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)
\(=\frac { 8\sqrt { 3 } +12-4\sqrt { 21 } }{ 12 } =\frac { 8\sqrt { 3 } }{ 12 } +\frac { 12 }{ 12 } -\frac { 4\sqrt { 21 } }{ 12 } \)
\(=\frac { 2\sqrt { 3 } }{ 3 } +1-\frac { \sqrt { 21 } }{ 3 } \)

First, draw a line segment AB = 5.26 units and extend it to C such that BC =1 unit. Let a be the mid-point of AC. Now,
draw a semi-circle with centre a and radius ac. Let us draw BD from point B, perpendicular to AC, which intersects the semi-circle at point D.
Hence, the distance BD represents \(\sqrt { 5.26 } \approx 2.29\) geometrivally.
Now, treat BC as the number line, draw an arc with centre B and radius BD, meeting AC produced at E. So, point E represents \(\sqrt { 5.26 } \) on the number line.

\(2{ y }^{ 3 }+y^{ 2 }-2y-1\)
Let  \(p(x)=2{ y }^{ 3 }+y^{ 2 }-2y-1\)
By trail, we find that
\(p(1)=2{ (1) }^{ 3 }+(1)^{ 2 }-2(1)-1\)
\(=2+1-2-1=0\)
\(\therefore\) By Factor Theorem, (y-1) is a factor of p(y)
Now,
\(2{ y }^{ 3 }+y^{ 2 }-2y-1\)
\(=2{ y }^{ 2 }(y-1)+3y(y-1)+1(y-1)\)
\(=(y-1)(2{ y }^{ 2 }+3y+1)\)
\(=(y-1)(2{ y }^{ 2 }+2y+y+1)\)
\(=(y-1)\{ 2{ y }(y+1)+1(y+1)\} \)
\(=(y-1)(y+1)(2y+1)\)

Let \(p(x)={ x }^{ 3 }-2{ x }^{ 2 }-x+2\)
By trail, we find that
\(p(1)={ (1) }^{ 3 }-2{ (1) }^{ 2 }-(1)+2\)
\(=1-2-1+2=0\)
\(\therefore \) By Factor Theorem, (x-1) is a factor of p(x).
Now,
\({ x }^{ 3 }-2{ x }^{ 2 }-x+2={ x }^{ 2 }(x-1)-x(x-1)-2(x-1)\)
\(=(x-1)({ x }^{ 2 }-x-2)\)
\(=(x-1)({ x }^{ 2 }-2x+x-2)\)
\(=(x-1)\{ x(x-2)+1(x-2)\} \)
\(=(x-1)(x-2)(x+1).\)

\(12x^2-7x+1\)
\(12x^{ 2 }-7x+1=12x^{ 2 }-4x-3x+1\)
\(=4x(3x-1)-1(3x-1)\)
\(=(3x-1)(4x-1)\)

\(5+2x\)
\(5+2x=0 \Rightarrow\ 2x=-5 \Rightarrow x=-\frac{5}{2}\)
\(\therefore\) Remainder
\(={ \left(- \frac { 5 }{ 2 } \right) }^{ 3 }+{ 3\left( -\frac { 5 }{ 2 } \right) }^{ 2 }+3{ \left(- \frac { 5 }{ 2 } \right) }+1\)
\(=-\frac{125}{8}+\frac{75}{15}-\frac{15}{2}+1=-\frac{27}{8}\)

Let \(p(x)=x^3+3x^2+3x+1\)
x+1
\(x+1=0 \quad\quad\Rightarrow x=-1\)
\(\therefore\) Remainder
\(=p(-1)=(-1)^3+3(-1)^2+3(-1)+1\)
\(=-1+3-3+1=0\)

\((i)\ { \left( { a }^{ b }+{ b }^{ a } \right) }^{ -1 }\)
\(\\ ={ \left( { 2 }^{ 3 }+{ 3 }^{ 2 } \right) }^{ -1 }={ \left( 8+9 \right) }^{ -1 }\)
\(\\ ={ \left( 17 \right) }^{ -1 }=\frac { 1 }{ 17 }\)
\( \\ (ii)\ { \left( { a }^{ b }+{ b }^{ b } \right) }^{ -1 }={ \left( { 2 }^{ 3 }+{ 3 }^{ 3 } \right) }^{ -1 }={ { \left( 4+27 \right) }^{ -1 } }\)
\(\\ ={ \left( 31 \right) }^{ -1 }=\frac { 1 }{ 31 } \)

0

8

\(\frac { 25 }{ 99 } \)

\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)=\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)\(\times \frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } +\sqrt { 6 } } \)
Multiplying and dividing by \(\sqrt { 7 } +\sqrt { 6 } \)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } -\sqrt { 6 } } =\sqrt { 7 } +\sqrt { 6 } \)

\({ \left( \sqrt { 5 } -\sqrt { 2 } \right) }{ \left( \sqrt { 5 } +\sqrt { 2 } \right) }={ \left( \sqrt { 5 } \right) }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 }\)
\(\\ =5-2=3\)

\((a)\ \sqrt { 3 } ,-\sqrt { 3 } \)
\(\\ (b)\ \sqrt { 3 } ,-\sqrt { 3 }\)
\( \\ (c)\ \sqrt { 3 } ,-\sqrt { 3 } \)