Polynomials

Polynomial, as 8 can be written as 8x⁰, and 0 is a whole number.

Polynomial, as the exponent of the variable in both term is a whole number.

Polynomial, as the exponent of the variable in both term is a whole number.

Polynomial, because the above equation can be re-written as , and the exponent of the variable in both term is a whole number.

Not a polynomial, as the exponent of the variable in is -1, i.e. not a whole number.

Not a polynomial, as the exponent of 'x' is not a whole number

Polynomial, as the exponent of the variable in both term is a whole number.

Not a polynomial, as the exponent of the variable in is -1, i.e. not a whole number.

False, A binomial can only have two terms, if a polynomial will have only one term then it will be a monomial.

False, Every binomial is a polynomial but every polynomial is not a binomial.

True, A binomial can only have two terms but there is no bound of degree of variable.

Example: x⁵- 32

False, it is not mandatory for example, zero of polymonial x - 1 is 1.

False, A quadratic equation i.e. polynomial with degree 2 can have two zeroes for example x² - 3x + 2 has two zeroes as 1 and 2.

True, As the terms with same degree will be added together.

(i) x² + x + 1 is a polynomial in one variable as it contains only one variable i.e. x.

(ii) y³ – 5y is a polynomial in one variable as it contains only one variable i.e. y

(iii) xy + yz + zx is a polynomial in three variable as it contains two variables i.e. x , y and z.

(iv) x² – 2xy + y² + 1 is a polynomial in two variables as it contains two variables i.e. x and y

(i) 2x – 1

The degree of polynomial 2x – 1 is one (1) because the highest power ( maximum exponent) of the variable x in the given expression is 1

(ii) –10

The degree of polynomial – 10 x⁰ is zero(0) because the highest power( maximum exponent) of the variable x in the given expression is zero (0).

(iii) x³ – 9x + 3x⁵

The degree of polynomial x³ – 9x + 3x⁵ is five (5) because the highest power( maximum exponent) of the variable x in the given expression is 5.

(iv) y³ (1 – y⁴)

The degree of polynomial y³ (1 – y⁴) is seven (7) because the highest power (maximum exponent) of the variable x in the given expression is 7

∵ y³ (1 – y⁴ ) = y³ – y⁷

The given polynomial is

(i) the degree of the polynomial is 6 as the highest power of the variable x in the given polynomial is 6.

(ii) the coefficient of x³ is 1/5 in the given polynomial as the given polynomial can be re-written as

(iii) the coefficient of x⁶ is – 1

(iv) the constant term is 1/5 as it has no variable x with it.

(i) The coefficient of x² in the polynomial + 𝑥²−1 is 1.

(ii) The coefficient of x² in the polynomial 3x – 5 is zero .

(iii) Let p(x) = (x – 1) (3x – 4)

= 3x² – 7x + 4

∴ The coefficient of x² in the polynomial 3x² – 7x + 4 is 3

(iv) The coefficient of x² in the polynomial is – 16

(2x – 5) (2x² – 3x + 1)

= 4x³ – 6x² – 10x² + 15x + 2x – 5

= 4x³ – 16x² + 17x – 5

The polynomial of the degree zero is constant, of degree one is linear , of degree two is quadratic and of degree three is cubic.

(i) 2 – x² + x³

It is a polynomial of the degree 3 so it is a cubic polynomial.

(ii) 3x³

It is a polynomial of the degree 3 so it is a cubic polynomial.

(iii) 5t – √7

It is a polynomial of the degree one(1) so it is a linear polynomial.

(iv) 4 – 5y²

It is a polynomial of the degree 2 so it is a quadratic polynomial.

(v) 3

It is a polynomial of the degree 0 so it is a constant polynomial.

(vi) 2 + x

It is a polynomial of the degree 1 so it is a linear polynomial.

(vii) y³ – y

It is a polynomial of the degree 3 so it is a cubic polynomial.

(viii) 1 + x + x²

It is a polynomial of the degree 2 so it is a quadratic polynomial.

(ix) t²

It is a polynomial of the degree 2 so it is a quadratic polynomial.

(x) √2x – 1

It is a polynomial of the degree 1 so it is a linear polynomial.

(i) an example of a monomial of degree 1 is 5x

(ii) an example of a monomial of degree y²⁰ + 9

(iii) trinomial of degree 2 is 1 + x + x²

let p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5

Putting x= 3 in given polynomial p(x)

P(3) = 3 × 3³ – 4 × 3² + 7 × 3 – 5 = 61

Similarly p( – 3) = 3 × – 3³ – 4 × ( – 3)² + 7 × ( – 3) – 5 = – 143

The value of given polynomial at x= 3 and – 3 is 61 and – 143 respectively.

given p(x) = 𝑥² – 4𝑥 + 3

Now p(2) = 2² – 4 × 2 + 3 = – 1

P( – 1) = ( – 1)² – 4 × ( – 1) + 3 = 8

P(1/2) = (1/2)² – 4 (1/2) + 3 = 1/4 – 2 + 3 = 5/4

∴ (2)−𝑝(−1) + 𝑝() = – 1 – 8 + 5/4 = – 31/4

(i) given p(x) = 10𝑥−4𝑥² –3

∴ p (0) = 10 × 0 – 4 × 0 – 3 = – 3

P (1) = 10 × 1 – 4 × 1 – 3 = 3

P ( – 2) = 10 × – 2 – 4 × ( – 2)² – 3 = – 39

The value of the polynomial (𝑥) = 10𝑥−4𝑥² –3 at p(0) , p(1) and p

( – 2) is – 3, 3 and – 39 respectively.

(ii) Given polynomial is p (y) = (y + 2) (y – 2)

= y² – 4

Now p (0) = 0² – 4 = – 4

P (1) = 1² – 4 = – 3

P ( – 2) = ( – 2)² – 4 = 0

Hence the value of the given polynomial at p(0) , p(1) and p

( – 2) is – 4, – 3 and 0.

(i) –3 is a zero of x – 3

False because zero of x – 3 = 3 ∵ x – 3 = 0 ⇒ x=3

(ii) is a zero of 3x + 1

True because zero of 3x + 1 = – 1/3 ∵ 3x + 1 = 0 ⇒ x = – 1/3

(iii) is a zero of 4 –5y

False because zero of 4 – 5y is 4/5 ∵ 4 – 5y =0 ⇒ – 5y = – 4 ⇒ y = 4/5

(iv) 0 and 2 are the zeroes of t² – 2t

T² – 2t = t(t – 2) = 0 ⇒ t = 0 or 2

Hence the zero of the polynomial are 0 and 2 . the given statement is true.

(v) –3 is a zero of y² + y – 6

y² + y – 6 = 0

⇒ y² + 3x – 2x – 6 = 0

⇒ y (y + 3) – 2(x + 3) = 0

⇒ (y – 2) (y + 3) =0

⇒ y = 2 or – 3

Hence – 3 is one of the zero of the polynomial. The given statement is true.

(i) p(x) = x – 4

For the zero of the polynomial p(x) we put p(x) = 0

⇒ x – 4= 0 ⇒ x = 4

Hence the zero of the polynomial is 4.

(ii) g(x) = 3 – 6x

For the zero of the polynomial Putting g(x) = 0 we get

3 – 6x = 0 ⇒ x = 3/6 = 1/2

Hence the zero of the polynomial is 1/2

(iii) q(x) = 2x –7

For the zero of the polynomial Putting q(x) = 0 we get

2x – 7 = 0 ⇒ x = 7/2

Hence the zero of the polynomial is 7/2

(iv) h (y) = 2y

For the zero of the polynomial Putting h(y)= 0 we get

2y =0 ⇒ y = 0

Hence the zero of the polynomial is 0

p(x) = (𝑥 –2)²−(𝑥 + 2)²

For the zero of the polynomial p(x) = 0

⇒ (x–2)²−(x + 2)² = 0

⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0 (a² – b² = (a – b) (a + b))

⇒ 2x ( – 4) = 0

⇒ – 8 x= 0

The zero of the polynomial are 0

The quotient is x³ + x² + x + 1 and the remainder is 2.

(i) Given p(x) = 𝑥³ – 2𝑥² – 4𝑥 – 1 and g(x) = x + 1

Here zero of g(x) = – 1

By using the remainder theorem

P(x) divided by g(x) = p( – 1)

P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0

So the remainder is 0

(ii) given p(𝑥) = 𝑥³ – 3𝑥² + 4𝑥 + 50, g(𝑥) = 𝑥 – 3

Here zero of g(x) = 3

By using the remainder theorem p(x) divided by g(x) = p(3)

p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62

hence the remainder is 62

(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1

Here zero of g(x) = 1/2

By using the remainder theorem p(x) divided by g(x) = p (1/2)

=

= 1/2 + 1 = 3/2

Hence the remainder is 3/2

(iv) p(𝑥) = 𝑥³ – 6𝑥² + 2𝑥 – 4, g(𝑥) = 1 – 3/2 𝑥

Here zero of g(x) = 2/3

By using the remainder theorem p(x) divided by g(x) = p(2/3)

Hence the remainder is – 136/27

here we would first find out the zero of g(x)and then put the value of in p(x) and solve it.

If p(a)=Q then p(x) is a multiple of g(x) and p(a) # Q then p(x) is not a multiple of g(x) where ‘a’ is a zero of g(x).

(i) g(x)=x – 2 (given)

then, zero of g(x) is 2

Now, p(2)=(2)³ – 5(2)² + 4(2) – 3

= 8 – 20 + 8 – 3 = – 7 ≠ 0

∴ p(x) is not the multiple of g(x) as the remainder ≠ 0.

(ii) p(x)= 2x³ – 11𝑥²− 4𝑥 + 5, g(x)= 2𝑥 + 1

Here the zero of g(x) = – 1/2

So p( – 1/2) = 2 × ( – 1/2)³ – 11 × ( – 1/2)² – 4 × ( – 1/2) + 5

Since the remainder is not equal to zero so p(x) is not a multiple of g(x).

(i) let p(x) = 69 + 11x −x² + x³ and g(x) = x + 3

We have to prove that g(x) is factor of p(x)

So the zero of g(x)= – 3

P( – 3) = 69 + 11( – 3) –( – 3)² + ( – 3) ³

= 69 – 69 = 0

Since the remainder is zero g(x) = x + 3 is factor of p(x)

(ii) Let p(x) = x + 2x³ – 9x² + 12 and g(x) =2x−3

We have to prove that g(x) is factor of p(x)

So the zero of g(x)= 3/2

Since the remainder is zero ∴ g(x) is factor of p(x)

(i) Let p(x) =3𝑥² + 6𝑥−24 and g(x) = x – 2

Zero of g(x) = 2

Here p(2) = 3(2)² + 6 (2) – 24 = 12 + 12 – 24 = 0

So g(x) is a factor of p(x)

(ii) Let p(x) = 4𝑥² + 𝑥−2 and g(x) = x – 2

Zero of g(x) = 2

P(2) = 4(2)² + 2−2 = 16 ≠ 0

Hence g(x) is not a factor of p(x)

let h (p) = 𝑝 ¹⁰ − 1,and g(p) = 𝑝 − 1

Putting g (p) = 0 ⟹ 𝑝 − 1 = 0 ⟹ 𝑝 = 1

According to factor theorem if g(p) is a factor of h(p) , then h(1) should be zero

h(1) = (1)¹⁰ − 1 = 1 − 1 = 0

⟹ g (p) is a factor of h(p).

Now, we have h (p) = 𝑝 ¹¹ − 1, g (p) = 𝑝 − 1

Putting g (p) = 0 ⟹ 𝑝 − 1 = 0 ⟹ 𝑝 = 1

According to factor theorem if g (p) is a factor of h (p) ,

Then h(1) = 0

⟹ (1)¹¹ – 1 = 0 hence g (p) the factor of h (p)

let p(x) = x ³ – 2mx² + 16, and g(x) = x + 2

Putting g(x) = 0 ⟹ x + 2 = 0 ⟹ x = – 2

According to factor theorem if p(x) is divisible by g(x), the remainder p(−2) should be zero.

p( – 2) = 0 ⟹ ( – 2)³ – 2m( – 2)² + 16 =0 – 8 – 8m + 16 = 0

⟹ 8m = 8 ⟹ m = 1

let g(x) = x + 2a and p(x) = x⁵ – 4a²x³ + 2x + 2a + 3

Putting g(x) = 0

⟹ x + 2a = 0 ⟹ x = – 2a

According to the factor theorem if g(x) is a factor of p(x) then p( – 2a) = 0

Now p ( – 2a) = ( – 2a)⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0

⟹ – 32a⁵ + 32a⁵ – 2a + 3 = 0 ⟹ – 2a = – 3

⟹

let p(x) = 8x⁴ + 4x³−16x² + 10x + m and g(x) = 2x – 1

Putting g(x) = 0 ⟹ 2x – 1 = 0 ⟹ x = 1/2

According to the factor theorem if g(x) is a factor of p(x) then p (1/2) = 0

p(1/2) = 8(1/2)⁴ + 4(1/2)³−16(1/2)² + 10(1/2) + 𝑚 = 0

⟹

let p(x) = ax³ + x² – 2x + 4a – 9 and g(x) = x + 1

Putting g(x) = 0 ⟹ x + 1= 0 ⟹ x = – 1

According to the factor theorem if g(x) is a factor of p(x) then p ( – 1) = 0

p( – 1) = a( – 1)³ + ( – 1)² – 2( – 1) + 4a – 9 = 0

⟹3a – 6 = 0 ⟹ a = 2

x² + 9x + 18

The third term “18” is factorized as: 3 and 6

Then,

x² + 3x + 6x + 18

⇒ x(x+ 3) + 6( x + 3)

⇒ (x+3)(x+6)

6x² + 7x – 3

The product of 1^st and the 3^rd term i.e. 18 can be factorized such that the sum or difference should be equal to the middle term.

Thus, the factors of 18 are 2 and 7

⇒ 6x² + 7x – 3

⇒ 6x² + 9x – 2x – 3

⇒ 3x(2+3) – 1(2x+3)

⇒ (3x – 1) (2x+3) Ans.

2x² – 7x – 15

The product of 1^st and the 3^rd term i.e. 30 can be factorized such that the sum or difference should be equal to the middle term.

Thus, the possible factors of 30 are 3 and 10

⇒ 2x² – 10x + 3x – 15

⇒ 2x(x – 5) +3(x – 5)

⇒ (2x+3) (x – 5)

84 – 2r – 2r²

= – 2(r² + r – 42)

= – 2 (r² + 7r – 6r – 42)

= – 2 {r(r + 7) – 6(r +7)}

let p(x) =2𝑥³ – 3𝑥² – 17𝑥 + 30

p(1) = 2 – 3 – 17 + 30 = 12 ≠ 0

p(2) = 2(2)³ – 3(2)² – 17(2) + 30 = 46 – 46 =0

hence x – 2 is a factor of p(x) 2x² + x – 15

p(x) = (x – 2) (2x² + x – 15)

= (x – 2) (2x² + 6x – 5x – 15)

= (x – 2) (2x (x + 3) – 5(x – 3))

= (x – 2) (2x – 5) (x + 3)

𝑥³ – 6𝑥² + 11𝑥 – 6

Let p(x) = 𝑥³ – 6𝑥² + 11𝑥 – 6

P(1) = 1³ – 6(1)² + 11(1) – 6 = 12 – 12= 0

⇒ (x – 1) is a factor of p(x)

p(x) = (x – 1) ( x² – 5x + 6)

= (x – 1) (x² – 3x – 2x + 6)

= (x – 1) (x(x – 3) – 2(x – 6))

= (x – 1) (x – 2) (x – 3)

𝑥³ + 𝑥² – 4𝑥 – 4

Let p(x) = 𝑥³ + 𝑥² – 4𝑥 – 4

p (1) = 1 + 1 – 4 – 4 ≠ 0

⇒ (x – 1) is not a factor of p(x)

p ( – 1) = ( – 1)³ + ( – 1)² – 4( – 1) – 4

⇒ – 1 + 1 + 4 – 4 = 0

⇒ x + 1 is the factor of p(x)

p(x) = (x + 1)( x – 2) (x + 2)

Let p(x) = 3𝑥³ – 𝑥² – 3𝑥 + 1

(1) = 3(1) ³ − (1)² − 3(1) + 1

= 3 − 1 − 3 + 1

= 4 − 4 = 0

⟹ (𝑥 − 1) is a factor of (𝑥).

p(x) = (x – 1) (3x² + 2x – 1)

= (x – 1) (3x² + 3x – 1x – 1)

(x – 1)( 3x(x + 1) – 1(x + 1))

= (x – 1) (3x – 1) (x + 1)

(i) (103)³

Given that: (103)³

= (100 + 3)³

= (100)³ + (3)³ + 3(100)²3 + 3(100)3²

[∵ (a + b)³ = (a³ + b³ + 3a² b + 3ab²)]

= 1000000 + 27 + 90000 + 2700

= 1092727

(ii). Given that: 101 × 102

= (100 + 1) × (100 + 2)

= (100)² + (1 + 2)100 + 1 × 2

[∵ (𝑥 + 𝑎)(𝑥 + 𝑏) = 𝑥² + (𝑎 + 𝑏)𝑥 + 𝑎 × 𝑏]

10000 + 300 + 2

= 10302

(iii). Given that: 999²

= (999)²

= (1000 − 1)²

= (1000)² − 2 × 1000 × 1 + 1²

[∵ (a – b)² = (a² – 2ab + b²) ]

= 1000000 − 2000 + 1

= 1000001 – 2000

= 998001

(i) 4x ² + 20x + 25 = (2x) ² + 2 × 2x × 5 + 5 ²

= (2x + 5)² [∵ 𝑎 ² + 2𝑎𝑏 + 𝑏 ² = (𝑎 + 𝑏)²]

(ii) Here p(x) = 9𝑦 ² − 66𝑦𝑧 + 121𝑧 ²

= (3𝑦) ² − 2 × 3𝑦 × 11𝑧 + (11𝑧) ² = (3x − 11z) ²

[∵ a² – 2ab + b² = (a – b)²]

(iii) Given that: (2𝑥 + 1 /3 ) ² − (𝑥 – 1/ 2 ) ²

[∵ 𝑎 ² − 𝑏 ² = (𝑎 − 𝑏) (𝑎 + 𝑏)]

=

=

(i) Given that: 9x² – 12x + 3

= 3(3x² – 4x + 1)

= 3(3x² – 3x − x + 1)

= 3[3x(x − 1) − 1(x − 1)]

= 3[(x − 1)(3x − 1)]

= 3(x − 1)(3x − 1)

(ii). Given that: 9x² – 12x + 4

= 9x² – 6x – 6x + 4

= 3x (3x − 2) − 2(3x − 2)

= (3x − 2) (3x − 2)

Given that: (4a − b + 2c) ²

= (4a) ² + (−b) ² + (2c)² + 2(4a)(−b) + 2(−b)(2c) + 2(2c)(4a)

[∵ (a + b + c) ² = a² + b² + c² + 2ab + 2bc + 2ca)]

= 16a² + b² + 4c² – 8ab – 4bc + 16ca

(ii). Given that: (3𝑎 − 5𝑏 − 𝑐)²

= (3a)² + (−5b)² + (−c)² + 2(3a)(−5b) + 2(−5b)(−c) + 2(−c)(3a)

[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)]

= 9a² + 25b² + c² – 30ab + 10bc – 6ca

(iii). Given that: (−𝑥 + 2𝑦 − 3𝑧)²

= (−𝑥)² + (2𝑦)² + (−3𝑧)² + 2(−𝑥)(2𝑦) + 2(2𝑦)(−3𝑧) + 2(−3𝑧)(−𝑥)

[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)]

= x² + y² + 9z² – 4xy – 12yz + 6zx

(i) Given that: 9x² + 4y² + 16z² + 12xy – 16yz – 24zx

=(3x² + (2y)² + (−4z)² + 2(3x)(2y) + 2(2y)(−4z) + 2(−4z)(3x)

= (3x + 2y – 4z)²

[∵ a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)²]

= (3x + y– 4z) (3x + 2y – 4z)

(ii). Given that: 25𝑥² + 16𝑦² + 4𝑧² − 40𝑥𝑦 + 16𝑦𝑧 − 20𝑧𝑥

=(−5x)² + (4y)² + (2z)² + 2(−5x)(4y) + 2(4y)(2z) + (2z)(−5x)

= (−5x + 4y + 2z)²

[∵ a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)²]

= (−5x + 4y + 2z)(−x + 4y + 2z)

(iii). Given that: 16x² + 4y² + 9z² – 16xy – 12yz + 24zx

= (4x)² + (−2y)² + (3z)² + 2(4x)(−y) + 2(−2y)(3z) + 2(3z)(4x)

= (4x – 2y + 3z)²

[∵ a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)² ]

= (4x– 2y + 3z)(4x – 2y + 3z)

Given that: a + b + c = 9 and 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 = 26

Using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca, we have(9)² = a² + b² + c² + 2(ab + bc + ca)

⟹ 81 = a² + b² + c² + 2(26)

⟹ 81 = a² + b² + c² + 52

⟹ 81 − 52 = a² + b² + c²

⟹ a² + b² + c²= 29

(i) Given that: (3a – 2b)³ = [3a + (−2b)]³

= (3a)³ + (−2b)³ + 3(3a)²(−2b) + 3(3a)(−2b)²

[∵ (a + b)³ = (a³ + b³ + 3a²b + 3ab²)]

= 27a³ – 8b³ – 54a2b + 36ab²

(ii). given

[∵ (a + b)³ =(a³ + b³ + 3a²b + 3ab²)]

=

(iii). Given that: (4 −13x)³

= (4)³ + (−13x)³ + 3(4)²(−13x) + 3(4) (−13x)²

[∵ (a + b)³ =(a³ + b³ + 3a²b + 3ab²)]

= 64 −127x³ −16x + 43x

(i) Given that: 1 – 64a³ – 12a + 48a²

= (1)³ + (−4a)³ + 3(1)²(−4a) + 3(1)(−4a)²

= (1 + (−4a))³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (1 – 4a)³

=

= (2p)³ + (15)³ + 3(2p)²(15) + 3(2p) (15)²

= (2p + 15)³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b) ³]

(i) Given that:

=

[∵ (a + b)(a² – ab + b²) = a³ + b³]

=

(ii) Given that: (x² − 1) (x⁴ + x² + 1)

= (x² − 1) [(x²)² + (x²) (1) + (1)²]

= (x²)³ − (1)³

[∵ (a – b) (a² + ab + b²) = a³ – b³]

= x⁶ − 1

(i) Given that: 1 + 64x³

= 1 + (4x)³

= (1 + 4x)[(1)² − (1)(4x) + (4x)²]

[∵ a³ + b³ = (a + b)(a² – ab + b²) ]

= (1 + 4x)(1 – 4x + 16x²)

(ii) Given that: a³ − 2√2b³

= (a)³ − (√2b)³

= (a − √2b) [(a)² + (a)(√2b) + (√2b)²]

[∵ a³ – b³ = (a + b)(a² + ab + b²) ]

= (a − √2b)(a² + √2ab + 2b²)

(2x − y + 3z)(4x² + y² + 9z² + 2xy + 3yz – 6xz)

= [2x + (−y) + 3z][(2x)² + (−y)² + (3z)²−(2x)(−y)−(−y)(3z) −(3z)(2x)]

= (2x)³ + (−y)³ + (3z)³ − 3(2x)(−y)(3z)

[∵ (a + b + c)(a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3abc ]

= 8x³ − y + 27z³ + 18xyz

(i) (a)³ + (−2b)³ + (−4c)³ − 3(a)(−2b)(−4c)

=[a + (−2b + ( – 4c)][(a)² + (−2b)² + (−4c)²−(a)(−2b)−(−2b)(−4c)− (−4c)(a)]

[∵a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ]

= (a– 2b – 4c)(a² + 4b² + 16c² + 2ab – 8bc + 4ca)

(ii). Given that: 2√2a³ + 8b³ – 27c³ + 18√2abc

= (√2a)³ + (2b)³ + (−3c)³ − 3(√2a)(2b)(3c)

=[√2a + (2b) + (−3c)][(√2a)² + (2b)² + (−3c)²−(√2a)(2b)−(2b)(−3c)− (−3c)(√2a)]

[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ]

= (√2a + 2b – 3c)(2a² + 4b² + 9c² − 2√2ab + 6bc + 3√2ca)

(i)

Let a =1/2, b=1/3and c= −5/6

=0/6

= 0

⟹ (1/2)³ + (1/3)³ + (5/6)³

= 3 (1/2) (1/3) (−5/6)

= −5/12

[∵a³ + b³ + c³–3abc =(a + b + c)(a² + b² + c² − ab− bc−ca)]

[ if 𝑎 + 𝑏 + 𝑐 = 0, 𝑎³ + 𝑏³ + 𝑐³ = 3𝑎𝑏𝑐]

(ii) . Given that: (0.2)³ − (0.3) ³ + (0.1)³

Let a = 0.2, b= −0.3 and c= 0.1

∴ a + b + c = 0.2 − 0.3 + 0.1 = 0.3 − 0.3 = 0

⟹ (0.2)³ + (−0.3)³ + (0.1)³

= 3(0.2)(−0.3)(0.1)

= −0.018

[∵ a³ + b³ + c³ – 3abc= (a + b + c)(a² + b² + c²− ab− bc− ca)]

[if a + b + c = 0, a³ + b³ + c³= 3abc]

Given that: (x– 2y)³ + (2y – 3z)³ + (3z − x)³

Let a = x – 2y, b= y− 3z and c= 3z − x

∴ a + b + c = x – 2y + 2y – 3z + 3z − x = 0

⟹ (x – 2y)³ + (2y – 3z)³ + (3z − x)³ =

3(x – 2y)(2y – 3z)(z − x)

[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c²− ab− bc− ca)]

[if a + b + c = 0, a + b³ + c³ = 3abc]

(i) Given that: 𝑥 + 𝑦 = −4

⟹ x + y + 4 = 0 … (i)

Now x³ + y³ – 12xy + 64

= x³ + y³ + 4³ − 3(x)(y)(4)

= (x + y + 4)[x² + y² + 4² − (x)(y) − (y)(4) − (4)(x)]

[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + 𝑏² + 𝑐² − ab− bc−ca)]

= (0)[x² + y² + 4² − (x)(y) − (y)(4) − (4)(x)]

[∵ x + y + 4 = 0, From (i)]

= 0

(ii). Given that: x = 2y + 6

⟹ x – 2y − 6 = 0 … (i)

Now x³ – 8y³ – 36xy − 216

= x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6)

= (x – 2y − 6)[x² + (−2y)2 + (−6)2−(x)(−2y)−(−2y)(−6)−(−6)(x)]

[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² − ab− bc− ca)]

= (0)[x² + (−2y)² + (−6)²−(x)(−2y)−(−2y)(−6)−(−6)(x)]

[∵ x − y− 6 = 0, From (i)]

= 0

Given

Area = 4a² + 4a − 3

= 4a² + 6a – 2a − 3

= 2a (a + 3) − 1(2a + 3)

= (2a + 3)(2a − 1)

⟹ Length = 2a + 3 and Breadth = 2a + 3