27p³-\(\frac { 1 }{ 216 } -\frac { 9 }{ 2 } { p }^{ 2 }+\frac { 1 }{ 4 } p\)
\(={ (3p) }^{ 3 }-{ \left( \frac { 1 }{ 6 } \right) }^{ 3 }-3.({ 3p) }^{ 2 }\frac { 1 }{ 6 } +3(3p){ \left( \frac { 1 }{ 6 } \right) }^{ 2 }\)
\(={ \left( 3p-\frac { 1 }{ 6 } \right) }^{ 3 }\)
\(=\left( 3p-\frac { 1 }{ 6 } \right) \left( 3p-\frac { 1 }{ 6 } \right) \left( 3p-\frac { 1 }{ 6 } \right) \)

\(\frac { 4 }{ 3\sqrt { 3 } -2\sqrt { 2 } } +\frac { 3 }{ 3\sqrt { 3 } -2\sqrt { 2 } } =\frac { 21\sqrt { 3 } +2\sqrt { 2 } }{ 19 } \)
\(=\frac { 21(1.732)+2(1.414) }{ 19 } \)
\(=\frac { 39.2 }{ 19 } \)
= 2.063

f(x) = x⁴ - 4x³ + 3x² - 2x + 1
f(0) = 1
f(-1)=(-1)⁴-4(-1)³+3(-1)²-2(-1)+1
= 1 + 4 + 3 + 2 + 1 = 11
f(2) = (2)⁴ - 4(2)³ + 3(2)² - 2(2) + 1
= 16 - 32 + 12 - 4 + 1
= 29 - 36 = -7
\(\therefore\) f(0) \(\times\) f(-1) = 11 \(\neq \) f(2)

(i)Infinitely many
(ii)x+y1=0, -x+y=5
(iii)(0,-1), (0, 5)

The given equation is x - 2y = 4
Put \(x=\sqrt{2}\), y=\(4\sqrt{2}\) in (1), we get
x-2y =\(\sqrt{2}-2(4\sqrt{2})\)
\(=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\) which is not 4.
\((\sqrt{2},\ 4\sqrt{2})\) which is not 4.
\((\sqrt{2}, 4\sqrt{2})\) is not a soluton of (1)

\(P\rightarrow I\)
\(\\ Q\rightarrow I\)
\(\\ R\rightarrow I\)
\(\\ S\rightarrow IV\)

Both the cross-streets are marked in the figure given.They are uniquely found because of the two reference lines we have used for locating them.

\((3x+4)(3x-5)\)
\((3x+4)(3x-5)=(3x+4)\{ 3x+(-5)\} \) | Using Identity IV
\({ =(3x) }^{ 2 }+\{ 4+(-5)\} (3x)+(4)(-5)\)
\(={ 9 }x^{ 2 }-3x-20\)

\(\therefore \ p(0)=2+0+2{ (0) }^{ 2 }-{ (0) }^{ 3 }=2\)
\(\\ p(1)=2+1+2{ (1) }^{ 2 }-{ (1) }^{ 3 }\)
\(\\ =2+1+2-1=4\)
and \(p(2)=2+2+2{ (2) }^{ 2 }-{ (2) }^{ 3 }\)
\(\\ =2+2+8-8=4\)

Let x=\(0.\overline { 47 } \) =0.4777....
Multiplying both sides by 10, we get
10x = 9.9999...
10x = 4.7777...
10x = 4.3+0.47777...
10x = 4.3+x
10x - x = 4.3
9x = 4.3
x = 4.3/9 = 43/90
Thus, \(0.\overline { 47 } \) = 43/90
Here p =4 3
q = 90(\(\neq 0\))

There can be infinitely many rational numbers between 3 and 4.
\(\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\\ \frac { 3+\frac { 7 }{ 2 } }{ 2 } =\frac { 13 }{ 4 } \)
\(\\ \frac { 3+\frac { 13 }{ 4 } }{ 2 } =\frac { 25 }{ 8 }\)
\( \\ \frac { 3+\frac { 25 }{ 8 } }{ 2 } =\frac { 49 }{ 16 } =\frac { 3+\frac { 49 }{ 16 } }{ 2 } =\frac { 97 }{ 32 } =\frac { 3+\frac { 97 }{ 32 } }{ 2 } =\frac { 193 }{ 64 } \)
Thus, six rational numbers between 3 and 4
\(\frac { 193 }{ 64 } ,\frac { 97 }{ 32 } ,\frac { 49 }{ 16 } ,\frac { 25 }{ 8 } ,\frac { 13 }{ 4 } \)and \(\frac { 7 }{ 2 } \)
Aliter
\(3=\frac { 3 }{ 1 } =\frac { 3\times 7 }{ 1\times 7 } =\frac { 21 }{ 7 } \)
\(\\ 4=\frac { 4 }{ 1 } =\frac { 4\times 7 }{ 1\times 7 } =\frac { 28 }{ 7 } \)
6 + 1 = 7
the six rational numbers between 3 and 4 can be taken as 
\(\frac { 22 }{ 7 } ,\frac { 23 }{ 7 } ,\frac { 24 }{ 7 } ,\frac { 25 }{ 7 } ,\frac { 26 }{ 7 } \) and \(\frac { 27 }{ 7 } \)

Substituting x=-1 and y=-1 in 9kx+12ky=63, we get
\(\Rightarrow\) 9k(-1)+12k(-1)=63
\(\Rightarrow\) -9k-12k=63
\(\Rightarrow\) -21k=63 k=-3

8-27a³-36a+54a²
=(2)³-(3a)³-18a(2-3a)
=(2)³-(3a)³-3\(\times\)2\(\times\)3a(2-3a)
=(2-3a)³.

(i)1
(ii)11

(0,3), (4, 6), (-4, 0)

(i) (0,-4)
(ii) (3,-5)

\(\left( 0,4\sqrt { 3 } \right) \)

(2,3), (2,0), (0,3)

IV

252

5

\(-\frac{7}{48}\)

(x-2)(2x-3)(3x-2)

\(\frac{5}{4}\)

Quotient =\(3x^3-x^3-x-1\)

Remainder=-5

trinomial

331/770, 166/385, 333/770

101/144, 47/72

\(a^7+ab^6\)
\(=a(a^6+b^6)\)

\(\frac{1}{y}=y^-1\) has negative exponent so, not a polynomial

\(x=\frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \)
\({ x }^{ 2 }={ \left[ \frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \right] }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 6-2\sqrt { 5 } } =\frac { 3+\sqrt { 5 } }{ 3-\sqrt { 5 } } \)
\(y=\frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \)
\({ y }^{ 2 }={ \left[ \frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \right] }^{ 2 }=\frac { 6-2\sqrt { 5 } }{ 6+2\sqrt { 5 } } =\frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } \)
\({ x }^{ 2 }+{ y }^{ 2 }=\frac { { \left( 3+\sqrt { 5 } \right) }^{ 2 }+{ \left( 3-\sqrt { 5 } \right) }^{ 2 } }{ \left( 3-\sqrt { 5 } \right) \left( 3+\sqrt { 5 } \right) } \)
\(=\frac { 9+5+6\sqrt { 5 } +9+5-6\sqrt { 5 } }{ 9-5 } \)
\(=\frac { 28 }{ 4 } \)
\({ x }^{ 2 }+{ y }^{ 2 }=7\)

Since total number of voters who do not cast therir cotes=40%
Hence total number of voters who cast their votes =60%
Let the total numbers of voters are x and number of voters who cast their votes is y.
Then according to the question
y=60% of \(x=\frac{60}{100}x\)
Now, when c=100, then
y=60
when x=200, then y=120
when c=300, then y=180

By plotting the points (100,60), (200,120) (300,180) on the graph and by joining them, we get the graph of equation (1) as shown in fig. From the graph, we see that
(i) If 300 voters cast their votes, then total number of votes=500

(ii) If the total number of voters are 1000, then number of votes cast=600.
(iii) Everyone should cast his vote to elect an honest candidate.

We have
y=4x+3
Table of solution

We plot the points (0,3) and (1, 7) on a graph paper and join the same by a ruler to get the line, which is the graph of the equation y = 4x + 3.
This gives the work wages graph of the given equation.
Now, on x-axis, take a point P(4, 0). From P draw a line parallel to y-axis intersecting the work wage graph at Q. From Q, draw a line parallel to x-axis to intersect the y-axis at R. We see that R is (0, 19).
Hence, the wages of a laborer who puts in 4 hours of work is Rs.19.

The given linear equation
\(\Rightarrow\ \ y={2\over3}x+{1\over3}\ \ \ \ . . . .(1)\)
Table of solution

We plot the points (1, 1) and (4, 3) on a graph paper and join the same by a ruler to get the line which is the graph of the equation \(y={2\over3}x+{1\over3}\)

From graph, we see that the point (7, 5) lies on the graph, so it is a solution of the linear equation.

(a) The coordinates A, B, C, D are (8,10), (-5,13), (13,5), (-4,-16) respectively.
(b) The triangle ABC has been shaded.

(i) 

(ii) see above figure
(iii) The figure obtained is a square.
(iv) Perimeter of the figure = 4 x 2 = 8 units
(v) Area of the figure = (2? = 4 square units
(vi) Join AC.
In right triangle ABC,
AC2 =AB2 +BC2 I By Pythagoras Theorem
= 2²+2²=4+4=8
\(\Rightarrow AC=\sqrt { 8 } =2\sqrt { 2 } \ units\)
Apala is correct.
So, the value 'Excellent Geometrical Knowledge is depicted by her finding.
(vii) The mathematical concept 'Coordinate Geometry' has been covered in this problem.

L.H.S=\((a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\)
\(=\left\{ (a+b)+(b+c)+(c+a) \right\} [(a+b)^2+(b+c)^2+(c+a)^2-(a+b)(b+c-(b_c)(c+a)-(c+a)(a+b)]\)
\(=2(a+b+c)[a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+c^2-ab-ac-b^2-bc-bc-ba-c^2-ca-ca-cb-a^2-ab]\)
Using Identity I
\(=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
\(=2(a^3+b^3+c^3-3abc)\)Using Identity VIII

\(\frac{87^3+13^3}{87^2-87\times13+13^3}\)
\(=\frac{(87+13(87^2-87\times13+12^2)}{(87^2-87\times13+13^2)}\)
= 87+13=100

\(\frac { 4 }{ { \left( 216 \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( 216 \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( 216 \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { \left( { 6 }^{ 3 } \right) }^{ -\frac { 2 }{ 3 } } } +\frac { 1 }{ { \left( { 4 }^{ 4 } \right) }^{ -\frac { 3 }{ 4 } } } +\frac { 2 }{ { \left( { 3 }^{ 5 } \right) }^{ -\frac { 1 }{ 5 } } } \)
\(=\frac { 4 }{ { 6 }^{ -2 } } +\frac { 1 }{ { 4 }^{ -3 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
\(=\frac { 4 }{ { 36 }^{ -1 } } +\frac { 4 }{ { 64 }^{ -1 } } +\frac { 2 }{ { 3 }^{ -1 } } \)
= 4 x 36 + 1 x 64 + 2 x 3
= 214