Since x + y + z = 0
∴ x + y = -z
or (x + y)³ = (-z)³
or x³ + y³ + 3xy(x + y) = -z³
or x³ + y³ + 3xy(-z) = -z³  [∵ x + y = (-z)]
or x³ + y³ - 3xyz = -z³
or (x³ + y³ + z³) - 3xyz = 0
or (x³ + y³ + z³) = 3xyz
Hence, if x + y + z = 0, then (x³ + y³ + z³) = 3xyz.

\(35y^2+13y-12\)
\(=35y^2+28y-15y-12\)
\(=7y(5y+4)-3(5y+4)\)
\(=(5y+4)(7y-3)\)
\(\because\)  The possible expressions for the length and breadth of the rectangle are 7y-3 and 5y+4

\((28)^3+(-15)^3+(-13)^3\)
=\(3(28)(-15)(-13)\)
\(\because (28)+(-15)+(-13)=0\)
= 16380

\((-12)^3+(7)^3+(5)^3\)
\((-12)^3+(7)^3+(5)^3=3(-12)(7)(5)\)
 \(\because (-12)+(7)+(5)=0\)
\(=-1260\)

L.H.S =\(x^3+y^3+z^3-3xyz\)
\(=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\)
Using Identity VIII
\(=\frac { 1 }{ 2 } (x+y+z)\left\{ 2(x^{ 2 }+y^{ 2 }+z^{ 2 }-xy-yz-zx) \right\} \)
\(=\frac { 1 }{ 2 } (x+y+z)(2x^2+2y^2-2xy-2yz-2zx)\)
\(=\frac { 1 }{ 2 } (x+y+z)\left\{ (x^{ 2 }-2xy+y^{ 2 })+(y^{ 2 }-2xy+z^{ 2 })+(z^{ 2 }-2zx+x^{ 2 }) \right\} \)
\(=\frac { 1 }{ 2 } (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]\)
Using Identity II

\(64m^3-343n^3\)
\(64m^3-343n^3=(4m)^3-(7n)^3\)
\(=(4m-7n)\left\{ (4m^{ 2 })(7n)+(7n)^{ 2 } \right\} \)
\(=(4m-7n)(16m^2+28mn+49n^2)\)

\(64a^3-27b^3-144a^2b+108ab^2\)
\(=(4a)^3-(3b)^3-3(4a)(3b)(4a-3b)\)
\(=(4a-3b)^3\) | Using Identity VII
\(=(4a-3b)(4a-3b)(4a-3b)\)

\(27+125a^3-135a+225a^2\)
\(27+125a^3-135a+225a^2\)
\(=(3)^3-(5a)^3-3(3)(5a)(3-5a)\)
\(=(3-5a)^3\)
\(=(3-5a)(3-5a)(3-5a)\)

\(8a^3+b^3+12a^2b+6ab^2\)
\(=(2a)^3+(b)^3+3(2a)(b)(2a+b)\)
\(=(2a+b)^3\) | Using Identify
\(=(2a+b)(2a+b)(2a+b)\)

\((2x-y+z)^2\)
\((2x-y+z)^{ 2 }={ \{ 2x+(-y)+z\} }^{ 2 }\)
\(={ (2x) }^{ 2 }+{ (-y) }^{ 2 }+{ (z) }^{ 2 }+2(2x)(-y)+2(-y)(z)+2(z)(2x)\) |Using Identity V
\(=4{ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }-4xy-2yz+4zx\) 

\(9x^2+6xy+y^2\)
\(9x^{ 2 }+6xy+y^{ 2 }=2(3x)(y)+{ (y) }^{ 2 }\)
 \(={ (3x+y) }^{ 2 }=(3x+y)(3x+y)\)  |Using Identity I

104 x 96
104 x 96=(100+4) x (100-4)
=(100)²-(4)²  |Using Identity III
=10000-16=9948

95 x 96
95 x 96=(90+5) x (90+6)
=(90)²+(5+6)(90)+(5)(6) | Using Identity IV
=8100+990+30=9120.
Aliter
95 x 96 = (100-5) x (100-4)
={100+(-5)} {100+(-4)}
=(100)²+{(-5)+(-4)}(100)+(-5)(-4) | Using Identity IV
=10000-900+20
=9120.

\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​\)
\(\left( { y }^{ 2 }+\frac { 3 }{ 2 } \right) \left( { y }^{ 2 }-\frac { 3 }{ 2 } \right) ​​=\left( z+\frac { 3 }{ 2 } \right) \left( z-\frac { 3 }{ 2 } \right) ​​\)| Where \(y^2=z\)
\(={ (z) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) | Using identity III
\(={ z }^{ 2 }-\frac { 9 }{ 4 } ={ ({ y }^{ 2 }) }^{ 2 }-\frac { 9 }{ 4 } \) | Substituting the value of z
\(={ y }^{ 4 }-\frac { 9 }{ 4 } .\)

\((x+8)(x-10)\)
\(​​​​(x+8)(x-10)=(x+8)\{ x+(-10)\} \)
\(={ x }^{ 2 }+[8+(-10)\} x+(8)(-10)\)  | Using Identity IV
\(={ x }^{ 2 }-2x-80\)

\(p(x)=k{ x }^{ 2 }-\sqrt { 2 } x+1\)
If x-1 is a factor of p(x), then p(1)=0 | By factor Theorem
\(\Rightarrow \ k{ (1) }^{ 2 }-\sqrt { 2 } (1)+1=0\)
\(\Rightarrow k-\sqrt { 2 } +1=0\)
\(\Rightarrow k=\sqrt { 2 } -1.\)

\(p(x)={ x }^{ 2 }+x+k\)
If x-1 is a factor of p(x), then p(1)=0 | By factor Theorem
\(\Rightarrow \ { (1) }^{ 2 }+(1)+k=0\)
\(\Rightarrow 1+1+k=0\)
\(\Rightarrow 1+1+k=0\)
\(\Rightarrow k=-2\)

\(x^4+3x^3+3x^2+x+1\)
Let \(p(x)=\)\(x^4+3x^3+3x^2+x+1\)
The zero of x+1 is -1.
\(​p(-1)=(-1)^4+3(-1)^3+(3-1)^2+(-1)+1\)
\(\\ =1-3+3-1+1=1\neq 0\)
\(\therefore\) By factors theorem, x+1 is not a factor of \(x^4+3x^3+3x^2+x+1\)

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

Let f (x) = 5x - 4x² + 3
Value of f(x) at x = 2
= f(2) = 5(2) - 4(2)² + 3
= 10 - 16 + 3 = -3   

Let f (x) = 5x - 4x² + 3
Value of f(x) at x = 0
= f (0) = 5 (0) - 4 (0)² + 3 = 3

We have 4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)
= [2x + (–y) + z]² (Using Identity V)
= (2x – y + z)² = (2x – y + z)(2x – y + z)
So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute (x + y)³. We have:
(x + y)³ = (x + y) (x + y)²
= (x + y)(x² + 2xy + y2)
= x(x² + 2xy + y²) + y(x² + 2xy + y²)
= x³ + 2x²y + xy² + x²y + 2xy² + y³
= x³ + 3x²y + 3xy² + y3
= x³ + y³ + 3xy(x + y)

(i) We know that
(x + y)³ = x³ + y³ + 3xy(x + y)
| Using Identity VI
⇒ x³ + y³ = (x + y)³ - 3xy(x + y)
⇒ x³ + y³ = (x + y){(x + y)² - 3xy)
⇒ x³ + y³ = (x + y)(x² + 2xy + y² - 3xy)
| Using Identity I
⇒ x³ + y³ = (x + y)(x² - xy + y²)
(ii) We know that
(x - y)³ = x³ - y³ - 3xy(x - y)
| Using Identity VII
⇒ x³ - y³ = (x - y)³ + 3xy(x - y)
x³ - y³ = (x - y){(x - y)² + 3xy}
⇒ x³ - y³ = (x - y)(x² - 2xy + y² + 3ry)
| Using Identity IV
⇒ x³ - y³ = (x - y)(x² + xy + y²).

(i) We have, 12x² - 7x + 1
= 12x² - (3 + 4)x + 1   [by splitting the middle term]
= 12x² - 3x - 4x + 1= 3x(4x - 1) - 1(4x - 1)
= (4x - 1)(3x - 1)
(ii) (x + 3)(2x + 1)
(iii) (3x - 2)(2x + 3)
(iv) (x + 1)(3x - 4)

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

(i) We have
(104)³ = (100 + 4)³
= (100)³ + (4)³ + 3(100)(4)(100 + 4)
(Using Identity VI)
= 1000000 + 64 + 124800
= 1124864
(ii) We have
(999)³ = (1000 – 1)³
= (1000)³ – (1)³ – 3(1000)(1)(1000 – 1)
(Using Identity VII)
= 1000000000 – 1 – 2997000
= 997002999

(By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.
So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.
So, 6x² + 17x + 5 = 6x² + (2 + 15)x + 5
= 6x² + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)

As x – 1 is a factor of p(x) = 4x³ + 3x² – 4x + k, p(1) = 0
Now, p(1) = 4(1)³ + 3(1)² – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3

Here, p(x) = x⁴ + x³ – 2x² + x + 1, and the zero of x – 1 is 1.
So, p(1) = (1)⁴ + (1)³ – 2(1)² + 1 + 1
= 2

Let p(x) = x² – 2x
Then p(2) = 2² – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x² – 2x.