\(\left(3^{\frac{1}{5}}\right)^{4}=3^{\frac{4}{5}}\)

Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of D ABD is given by
\(s=\frac{100+100+160}{2} \mathrm{~m}=180 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ABD}=\sqrt{180(180-100)(180-100)(180-160)}\)
\(=\sqrt{180 \times 80 \times 80 \times 20} \mathrm{~m}^{2}=4800 \mathrm{~m}^{2}\)
Therefore, each of them will get an area of 4800 m².

Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig.).

Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)
Therefore, 15x = 300, which gives x = 20.
So the sides of the triangle are 3 x 20 m, 5 x 20 m and 7 x 20 m
i.e., 60 m, 100 m and 140 m.
We have s \(=\frac{60+100+140}{2} \mathrm{~m}=150 \mathrm{~m}\)
and area will be \(\sqrt{150(150-60)(150-100)(150-140)} \mathrm{m}^{2}\)
\(=\sqrt{150 \times 90 \times 50 \times 10} \mathrm{~m}^{2}\)
\(=1500 \sqrt{3} \mathrm{~m}^{2}\)

\(18\sqrt { 21 } \) cm²

\(25\sqrt { 3 } \) cm²

\((5x-3y+z)(25x^2+9y^2+z^2+15zy+3yz-5zx)\)

\(2(2x-y)(4x^2+2xy+y^2)\)

Using Identity V, we have
(4a – 2b – 3c)² = [4a + (–2b) + (–3c)]²
= (4a)² + (–2b)² + (–3c)² + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)
= 16a² + 4b² + 9c² – 16ab + 12bc – 24ac

So, we find that the remainder is 0.
Here p(x) = x³ + 1, and the root of x + 1 = 0 is x = –1. We see that
p(–1) = (–1)³ + 1
= –1 + 1
= 0

We want to write \(\frac { 1 }{ \sqrt { 2 } } \)as an equivalent expression in which the denominator is a rational number. We know that \(\sqrt{2} \cdot \sqrt{2} \) is rational. We also know that multiplying \(\frac{1}{\sqrt{2}} \text { by } \frac{\sqrt{2}}{\sqrt{2}}\) will give us an equivalent expression, since \(\frac{\sqrt{2}}{\sqrt{2}}=1\) . So, we put these two facts together to get 
\(\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
In this form, it is easy to locate \(\frac{1}{\sqrt{2}}\) on the number line. It is half way between 0 and \(\sqrt{2} \text {. }\)

Smallest side = \(\frac { 12 }{ 25+17+12 } \times 540\) = 120 m.

Perimeter = 17+33+20 = 70 cm

\(36=\frac { 9\times Perpendicular }{ 2 } \)
\(\quad \Rightarrow \)  Perpendicular = 8 cm.

\((x-\frac{1}{x})(x+\frac{1}{x})(x^2+\frac{1}{x^2})\)
\((x^2-\frac{1}{x^2})(x^2+\frac{1}{x^2})=x^4-\frac{1}{x^4}\)

\(4a^2+9b^2+c^2+12ab+4ac+6bc\)
\(=(2a)^2+(3b)^2+(c)^2+2(2a)(3b)+2(3c)(c)+2(c)(2a)\)
\((2a+3b+c)^2\)

\(x^2+2 x+3=0\)
\(\Rightarrow\quad x=\frac{-2\pm\sqrt{4-12}}{2}=\frac{-2\pm2\sqrt{2}i}{2}\)
\(=-1\pm\sqrt{2}i\)

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x + 3y - 4z)² [Using Identity V]
= (2x + 3y - 4z)(2x + 3y - 4z)
(ii) 2x² + y²+ 8z² - 2 \(\sqrt{2}\)xy + 4\(\sqrt{2}\) yz - 8xz
= (-2 \(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2 (-\(\sqrt{2}\)x)(y) + 2 (2\(\sqrt{2}\)z)(y) + 2 (2\(\sqrt{2}\)z) (-\(\sqrt{2}\)x)
= (-\(\sqrt{2}\)x + y + 2 \(\sqrt{2}\))²
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)( -\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)

Let ABCD is a rhombus shaped field whose each side is 30 m and longer diagonal AC is 48 m. Clearly, the diagonal AC divides the rhombus into two triangles, ΔABC and ΔADC which are congruent.
ஃ  Area of ΔABC = Area of ΔADC
Also, ΔABC and ΔADC have equal perimeters.

Now, semi-perimeter of ΔABC,
\(s=\frac{a+b+c}{2}=\frac{30+30+48}{2}=\frac{108}{2}=54m\)
ஃ Area of ΔABC 
\(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{54\times(54-30)(54\times30)(54-48)}\)
\(=\sqrt{54\times24\times24\times6}\)
\(=\sqrt{18\times3\times24\times24\times6}\)
\(=18\times27\times=432m^2\)
ஃ Area of rhombus ABCD = 2 x Area of ΔABC
\(=2\times432=864m^2\)
Now, each cow will get =\(\frac{864}{18}=48m^2\) area.
Hence, each cow will get 48 m²  area.

\(-\frac { 31 }{ 4 } \)

Given, \(=\frac { 1 }{ 7-4\sqrt { 3 } } =\frac { 1 }{ 7-4\sqrt { 3 } } \times \frac { 7+4\sqrt { 3 } }{ 7+4\sqrt { 3 } } \)
\(=\frac { 7+4\sqrt { 3 } }{ { (7) }^{ 2 }-{ (4\sqrt { 3 } ) }^{ 2 } } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { 7+4\sqrt { 3 } }{ 49-48 } =7+4\sqrt { 3 } \)
\(and\ b=\frac { 1 }{ 7+4\sqrt { 3 } } \frac { 1 }{ 7-4\sqrt { 3 } } \times \frac { 7-4\sqrt { 3 } }{ 7-4\sqrt { 3 } } \)
\(=\frac { 7-4\sqrt { 3 } }{ { (7) }^{ 2 }-{ (4\sqrt { 3 } ) }^{ 2 } } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(=\frac { 74\sqrt { 3 } }{ 49-48 } =7-4\sqrt { 3 } \)
\(\therefore \ a+b=7+4\sqrt { 3 } +7-4\sqrt { 3 } =14\)
and \(ab=(7+4\sqrt { 3 } )(7-4\sqrt { 3 } )\)
\(={ (7) }^{ 2 }-{ (4\sqrt { 3 } ) }^{ 2 }=49-48\quad \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
= 1
(i) \(\because \ { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab\)
\(\Rightarrow \ { (14) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2\)
\(\therefore \ { a }^{ 2 }+{ b }^{ 2 }=196-2\)
= 194
(ii) \(\because \ { (a+b) }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+3ab(a+b)\)
\({ a }^{ 3 }+{ b }^{ 3 }={ (a+b) }^{ 3 }-3ab(a+b)\)
\(\Rightarrow ={ (14) }^{ 3 }-3\times 1\times 14\)
= 2744 - 42
= 2702

(i) We have, \(\frac { 1 }{ \sqrt { 7 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } \) , we get
\(\frac { 1 }{ \sqrt { 7 } } \times \frac { \sqrt { 7 } }{ \sqrt { 7 } } =\frac { \sqrt { 7 } }{ 7 } \)
(ii) We have,\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } +\sqrt { 6 } ,\) we get
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \left( \sqrt { 7 } +\sqrt { 6 } \right) }{ \left( \sqrt { 7 } +\sqrt { 6 } \right) } \)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ { \left( \sqrt { 7 } \right) }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 } } [\because (a-b)(a+b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 1 } \)
\(=\sqrt { 7 } +\sqrt { 6 } \)
(iii) \(\left[ \frac { \sqrt { 5 } -\sqrt { 2 } }{ 3 } \right] \)
(iv) \(\left[ \frac { \sqrt { 7 } +1 }{ 3 } \right] \)