Now, for n = - 2
5n² + 5n-2 = 8 and n³ = (- 2)³ = (- 2) x (- 2) x (- 2) = - 8
Combining,
n³ + 5n² + 5n - 2 = - 8 + 8 = o.

In 5n² + 5n - 2, we have
For n = - 2, 5n - 2 = - 12 and,5n² = 5 x (- 2)² = 5 x 4 = 20
Combining,
5n² + 5n - 2 = 20 - 12 = 8

Value of a³ + a²b + ab²+b³at a = 1 and b = - 2
= (1)³+ (1)²(-2) + (1)(- 2)² + (-2)³
=1-2+4-8
= 5-10
=-5

Value of a² + b² + 3ab at a = 1 and b = - 2
= (1)² + (-2)² + 3(1)(- 2)
=1+4-6
=5-6
= -1

- 3x³y² +2x²y³
             -3x²y³-5y^{4
________________}
- 3x³y² +x²y³-5y^{4
_________________}
Sum =- 3x³y² +x²y³-5y⁴
Now,                 x⁴ +x³y² + x²y³ + y⁴
                            - 3x³y² - x²y³ - 5y⁴
                            (+)     (+)        (+)       
                       __________________ 
Difference =     x⁴ +4x³y²+2x²y³+ 6y⁴
                       ___________________

A = 3x² + 2x and B = 3x + 1
A - B = 3x² + 2x - 3x - 1
= 3x² -x-1

m+n
m-n
Adding, 2m + 0 = 2m

mn + 5-2
mn + 3
Adding, 2mn + 8-2 = 2mn + 6

2x²+ x + 7 + 4(x - 5) = 2x²+ x + 7 + 4x - 20
= 2x²+ x + 4x + 7-20
= 2x²+ 5x-13;
If x = 3, given expression
= 2(3)²+ 5(3)- 13
= 18 + 15 - 13 = 20

The number z multiplied by itself = z X z = z².

The required expression is obtained by subtracting (4b² - a²) from (b² - 4a²).
∴ The required expression
= (b² - 4a²) - (4b² - a²)
= b² - 4a² - 4b² + a²
= (b² _ 4b²) + (-4a² + a²)
= (1 - 4)b² + (-4 + 1)a²
= (-3)b² + (-3)a²
= -3b² - 3a²
Thus, the required expression is (-3b² - 3a²).

The required expression is obtained by
subtracting (4a² - b + 2) from (3a² - 3b + 6).
∴ We have (3a² - 3b + 6) - (4a² - b + 2)
= 3a² - 3b + 6 - 4a² + b - 2
= (3a² - 4a²) + (-3b + b) + (6 - 2)
= (3 - 4)a² + (-3 + l)b + (4)
= -a² - 2b + 4
Thus, the expression (-a² - 2b + 4) can be subtracted from 3a² - 3b + 6 to get 4a² - b + 2.

Sum of 5x² - 6x + 4 and -x²r - 2x + 3
= (5x² - 6x + 4) + (-4x² - 2x + 3)
= 5x² - 6x + 4 - 4x² - 2x + 3
= (5x² - 4x²) + (-6x - 2x) + (4 + 3)
= (5 - 4)x² + (-6 - 2)x + (7)
= (I)x² + (-8x) + 7
=x²-8x+7
Now subtract x² - 8x + 7 from 0, we get 
∴  0 - [x² - 8x + 7] = 0 - x² + 8x - 7
= -x² + 8x - 7

The required expression
= (2x³-3x² + 4x + 5) - (2x³ + 7x² - 2x + 7)
= 2x³ - 3x² + 4x + 5 - 2x³ - 7x² + 2x - 7
= (2x³ - 2x³) + (-3x²-7x²) + (4x + 2x)+ (5-7)
= (2 - 2)x³ + (-3 - 7)x² + (4 + 2)x + (5 - 7)
= (O)x³ + (-10)x² + (6)x + (-2)
=-10x²+6x-2
Thus, the expression (2x³ - 3x² + 4x + 5) is greater than ( 2x³ + 7x² - 2x + 7) by the expression (-10x² + 6x - 2).

A-B = (3c² - 4b - 1) - (6c²+ 3b - 8)
= 3a² - 4b - 1 - 6c² - 3b + 8
= (3a² - 6c²) + (-4b - 3b) + (-1 + 8)
= (3 - 6)a² + (-4 - 3)b + (-1 + 8)
= (-3)a² + (-7)b + (7)
= -3a² - 7b + 7
∴ A - B + C = (-3a² - 7b + 7) + (4a² - 9b + 3)
= -3a² - 7b + 7 + 4a² - 9b + 3
= (-3a² + 4a²) + (-7b - 9b) + (7 + 3)
= (-3 + 4)a² + (-7 - 9)b + (10)
= (1)a² + (-16)b + 10
= a² - 16b + 10

(a)Total amount deposited in account = (5x² + 3x-2) + (x²-x + 5)
=5x² + 3x-2+ x²-x + 5
=5x² + x²+ 3x-x-2+ 5
=6x²+2x+3
Balance amount
= (6x²+2x+3)-(2x²-x-3)
= 6x²+2x+3-2x²-x-3
= 6x²-2x²+ 2x + x + 3 + 3
=4x²+3x+6.
(b) Addition of algebraic expressions.
(c) Value: Economy is everywhere.

\(({1\over3}y^2-{4\over7}y^2+5)-({2\over7}y-{2\over3}y^3+2)={1\over3}y^2-{4\over7}y+5-{2\over7}y+{2\over3}y^2-2\)
\(=({1\over3}+{2\over3})y^3-({4\over7}+{2\over7})y+5-2\)
\(={3\over3}y^2-{6\over7}y+3\)
\(=y^2-{6\over7}y+3\)
\(=(2)^2-{6\over7}(2)+3\)
\(=4-{12\over7}+3\)
\(={28-12+21\over7} \)                        \([\because y=2]\)
\(={37\over7}\).

Sum of (8a - 6a² + 9) and (-10a - 8 + 8a²)
= (8a - 6a² + 9) + (- 10a - 8 + 8a²)
= 8a-6a² + 9-10a-8 + 8a²
= (8a²- 6a²)+ (8a -10a) + (9 - 8)
= 2a²-2a + 1
Subtraction of (2a² - 2a + 1) from - 3
= -3-(2a²-2a + 1)
=-2a²+2a-1-3
= -2a² + 2a-4

\((3x^2-{1\over5}x+{7\over3})+(-{1\over4}x^2+{1\over3}x-{1\over6})+(-2x^2-{1\over2}x+5)\)
=\(3x^2-{1\over4}x^2-2x^2-{1\over5}x+{1\over3}x-{1\over2}x+{7\over3}-{1\over6}+5\)
\(=(3-{1\over4}-x^2)x^2+(-{1\over5}+{1\over3}-{1\over2})x+({7\over3}-{1\over6}+5)\)
\(=({12-1-8\over 4})x^3+({-6+10-15\over30})x+({14-1+30\over6})\)
\(={3\over4}x^2-{11\over30}x+{43\over6}\)

We have \(({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4})\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
= \({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4}\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
 =\(({1\over5}+{2\over3})x^3+(-{3\over2}-{8\over5})x^2+({2\over3}-{3\over2})x+({1\over4}+1)\)
\(={13x^3\over 15}-{31x^2\over10}-{5x\over 6}+{5\over4}\) 

Expression mn² + m²n + 8mn + 9
Term I :mn²
degree = 1 + 2 = 3
Term II :m²n
degree = 1 + 2 = 3
Term III : 8mn
degree = 0 + 1 + 1 = 2
Term IV: 9
degree = 0

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

A = 2 + 4x + 8x²
B = -3-5x + x²
C = 1 + 3x-7x²
A+B+C=?
A + B + C = (2 + 4x + 8x²) + (- 3 - 5x + x²)+ (1 + 3x-7x²)
= 2 + 4x + 8x² - 3 - 5x + x² + 1 + 3x -7x²
= x²(8 + 1 - 7) + x( 4 - 5 + 3) + 2 - 3 + 1
=2x²+2x+0
=2x²+2x
= 2x(x + 1).

A = 2a - 3b
B=-3a+4b
C = -a + b
A+B+C=?
A + B + C = (2a-3b) + (-3a + 4b) + (-a + b)
= 2a-3b-3a + 4b-a + b
= a(2 - 3 -1) + b(- 3 + 4 + 1)
= a(2 - 4) + b(1 + 1)
=-2a+2b
A+B-C=?
A + B - C = (2a- 3b) + (- 3a + 4b) - (- a + b)
= 2a-3b-3a + 4b + a-b
= a(2-3 + 1) + b(-3 + 4-1)
= a(0) + b(-4 + 4)
= a(0) + b(0)
= 0.