Factorisation

Given, f(x)= x³ – 3x +2

In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0

f(1)=(1)³−3.(1)+2=0

We observe that f(1) = 0

From factor theorem, we can say, (x−1) is a factor of f(x)

x³ – 3x +2 = x³ – x² + x² − x − 2x +2

= x²(x−1)+x(x−1)−2(x−1)

= (x−1)(x²+x−2)

= (x−1)(x²+2x−x−2)

= (x−1)( x(x+2) – (x+2) )

= (x−1)(x−1)(x+2)

= (x−1)²(x+2)

Given, f(x)= x³ + 2x + 3

In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0

Since, each term of f(x) is positive here; so for positive value of x we shall not get the value of f(x) as zero.

Hence, for the negative value of x, the value of f(x) can be zero.

f(−1)=(−1)³+2.(−1)+3=0

We observe that f(−1) = 0

From factor theorem, we can say, (x+1) is a factor of f(x)

x³ + 2x + 3 = x³ + x² – x² – x + 3x + 3

= x²(x+1)−x(x+1)+3(x+1)

= (x+1)(x²−x+3)

Given, f(a) = a³ – 12a – 16

In f(a) putting a=±1, ±2, ±3, we see for which value of a, f(a)=0

f(−2)=(−2)³−12.(−2)−16=0

We observe that f(−2) = 0

From factor theorem, we can say, (a+2) is a factor of f(a)

a³ – 12a – 16 = a³ – 12a – 16

= a³+2a²–2a²– 4a – 8a − 16

= a²(a+2)−2a(a+2)−8(a+2)

= (a+2)(a²−2a−8)

= (a+2)(a²−4a+2a−8)

= (a+2)( a(a−4) + 2(a−4) )

= (a+2)(a−4)(a+2)

= (a+2)²(a−4)

Given, f(x) = x³ – 6x + 4

In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0

f(2)=(2)³−6.2+4=0

We observe that f(2) = 0

From factor theorem, we can say, (x−2) is a factor of f(x)

x³ – 6x + 4 = x³ – 6x + 4

= x³ −2x² +2x² −4x −2x +4

= x²(x−2)+2x(x−2)−2(x−2)

= (x−2)(x²+2x−2)

Given, f(x)= x³ – 19x – 30

In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0

f(−2)=(−2)³−19.(−2)+30=0

We observe that f(−2) = 0

From factor theorem, we can say, (x+2) is a factor of f(x)

x³ – 19x – 30 = x³ – 19x – 30

= x³+2x²−2x²−4x−15x −30

= x²(x+2)−2x(x+2)−15(x+2)

= (x+2)(x²−2x−15)

= (x+2)(x²−5x+3x−15)

= (x+2)( x(x−5) + 3(x−5) )

= (x+2)(x−5)(x+3)

Given, f(a) = 4a³ – 9a² + 3a + 2

In f(a) putting a=±1, ±2, ±3, we see for which value of a, f(a)=0

f(1)=4.(1)³ −9.(1)² +3.(1)+2=0

We observe that f(1) = 0

From factor theorem, we can say, (a−1) is a factor of f(a)

4a³–9a²+3a+2= 4a³−4a²−5a²+5a−2a+ 2

= 4a²(a−1)−5a(a−1)−2(a−1)

= (a−1)(4a²−5a−2)

Given, f(x)= x³ – 9x² + 23x – 15

In f(x) putting x=±1, ±2, ±3, we see for which value of x, f(x)=0

f(1)=4.(1)³ −9.(1)² +3.(1)+2=0

We observe that f(1) = 0

From factor theorem, we can say, (x−1) is a factor of f(x)

x³ – 9x² + 23x – 15= x³ – 9x² + 23x – 15

= x³−x²−8x²+8x+15x−15

= x²(x−1)−8x(x−1)+15(x−1)

= (x−1)(x²−8x+15)

= (x−1)(x²−5x−3x+15)

= (x−1)( x(x−5) −3(x−5) )

= (x−1)(x−5)(x−3)

Given, f(a)= 5a³ + 11a² + 4a –2

In f(a) putting a=±1, ±2, ±3, we see for which value of a, f(a)=0

f(−1)=5.(−1)³+11.(−1)²+4.(−1)−2=0

We observe that f(−1) = 0

From factor theorem, we can say, (a+1) is a factor of f(a)

5a³ + 11a² + 4a –2 = 5a³ + 11a² + 4a –2

= 5a³+5a²+6a²+6a−2a−2

= 5a²(a+1)+6a(a+1)−2(a+1)

= (a+1)(5a²+6a−2)

Given, f(x)= 2x³ – x² + 9x + 5

In f(x) putting x=±1, , ±2, ±3, we see for which value of x, f(x)=0

f(−)=2.(− )³−(−)²−9.(− )+5=0

We observe that f(−) = 0

From factor theorem, we can say, for (x=−), (2x+1) is a factor of f(x)

2x³ – x² + 9x + 5 = 2x³ – x² + 9x + 5

= 2x³+x²−2x²− x + 10x +5
= x²(2x+1)−x(2x+1)+5(2x+1)

= (2x+1)(x²−x+5)

Given, f(y)= 2y³ – 5y² –19y+ 42

In f(y) putting y=±1, ±2, ±3, we see for which value of y, f(y)=0

f(2)=2.(2)³−5.(2)²−19.2+42=0

We observe that f(2) = 0

From factor theorem, we can say, (y−2) is a factor of f(y)

2y³ – 5y² –19y+ 42= 2y³ – 5y² –19y+ 42

= 2y³−4y²−y²+2y−21y+42

= 2y²(y−2)−y(y−2)−21(y−2)

= (y−2)(2y²−y−21)

Given,

⇒ This can be written as

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

Given, m² + + 2 – 2m –

Given, 9p² – 24pq + 16q² + 3ap – 4aq

⇒ (3p)² – 24pq + (4q)² + 3ap – 4aq

⇒ (3p)² – 2(3p)(4q) + (4q)² + 3ap – 4aq

[from Identity I, (x – y)² = x² – 2xy + y²]

⇒ (3p – 4q)² + a(3p – 4q)

⇒ (3p – 4q)[(3p – 4q) + a]

⇒ (3p – 4q)(a + (3p – 4q))

Given, 4x⁴ + 81

⇒ 4x⁴ + 3⁴

⇒ (2x²)² + 9²

⇒ (2x² + 9)² – 2(2x²)(9)

⇒ (2x² + 9)² – 36x²

⇒ (2x² + 9)² – (6x)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((2x² + 9) – (6x)) ((2x² + 9) + (6x))

⇒ (2x² + 9 – 6x)(2x² + 9 + 6x)

Given, x⁴ – 7x² + 1

Can be written as

⇒ (x² + 1)² – 9x²

⇒ (x² + 1)² – (3x)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((x² + 1) – 3x)((x² + 1) + 3x)

⇒ (x² – 3x + 1)(x² + 3x + 1)

Given, p⁴ – 11p²q² + q⁴

⇒ (p²)² – 11p²q² + (q²)²

⇒ (p²)² – 2p²q² – 9 p²q² + (q²)²

⇒ (p²)² – 2p²q² + (q²)² – 9 p²q²

⇒ (p² – q²)² – (3pq)²

⇒ ((p² – q²) – 3pq)((p² – q²) + 3pq)

⇒ (p² – q² – 3pq)(p² – q² + 3pq)

Given, a² + b² – c² – 2ab

⇒ This can be written as a² – 2ab + b² – c²

⇒ From the identity II, a² – 2ab + b² = (a – b)²

∴ (a – b)² – c²

⇒ ((a – b) – c)((a – b) + c)

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (a – b – c)(a – b + c)

Given, 3a(3a + 2c) – 4b(b + c)

⇒ 9a² + 6ac – 4b² – 4bc

⇒ (3a)² + 2(3)(ac) – (2b)² – 4bc + c² – c²

⇒ (3a)² + 2(3)(ac) + c² – ((2b)² + 2(2b)(c) + c²)

⇒ (3a + c)² – (2b + c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (3a + c – 2b – c)(3a + c + 2b + c)

⇒ (3a – 2b)(3a + 2b + 2c)

Given, a² – 6ab + 12bc – 4c²

⇒ a² – 4c² – 6ab + 12bc

⇒ (a² – (2c)²) – 6b(a – 2c)

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (a + 2c)(a – 2c) – 6b(a – 2c)

⇒ (a – 2c)(a + 2c – 6b)

Given,3a² + 4ab + b² – 2ac – c²

⇒ 3a² + 4ab + b² – 2ac – c²

⇒ 4a² + 4ab + b² – a² – 2ac – c²

⇒ From the identity I, a² + 2ab + b² = (a + b)²

⇒ 4a² + 4ab + b² – (a² + 2ac + c²)

⇒ (2a + b)² – (a + c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((2a + b) – (a + c))

((2a + b) + (a + c)

⇒ (2a + b – a – c)(2a + b + a + c)

Given,x² – y² – 2(3ax) + 2ay + 9a² – a²

⇒ x² – 2(3a)(x) + (3a)² – (a² – 2ay + y²)

⇒ (x – 3a)² – (a – y)²

⇒ (x – 3a – a + y)(x – 3a + a – y)

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (x – 4a + y)(x – 2a – y)

Given, a² – 9b² + 4c² – 25d² – 4ac + 30bd

⇒ a² – 9b² + 4c² – 25d² – 4ac + 30bd

⇒ a² – 4ac + 4c² – 9b² – 25d² + 30bd

⇒ a² – 2(2ac) + (2c)² – (3b)² – (5d)² + 2(3b)(5d)

⇒ (a – 2c)² – (3b – 5d)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ ((a – 2c) – (3b – 5d))(a – 2c – 3b + 5d)

⇒ (a – 2c – 3b – 5d)(a – 2c – 3b + 5d)

Given, 3a² – b² – c² + 2ab – 2bc + 2ca

⇒ 4a² – a² – b² – c² + 2ab – 2bc + 2ca

⇒ (2a)² – (a² + b² + c² – 2ab + 2bc – 2ca)

⇒ (2a)² – (a – b – c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (2a – (a – b – c))(2a + a + b + c)

⇒ (a + b + c)(3a + b + c)

Given, x² – 2x – 22499

⇒ x² – 151x + 149x – 22499

⇒ x(x – 151) + 149(x – 151)

⇒ (x + 149)(x – 151)

Given, (x² – y²)(a² – b²) + 4abxy

⇒ ((x – y)(x + y)(a – b)(a + b)) + 4abxy

⇒ (ax)² – (bx)² – (ay)² + (by)² + 4abxy

⇒ (ax)² + (by)² + 4abxy – (bx)² – (ay)²

⇒ (ax)² + (by)² + 2abxy + 2abxy – (bx)² – (ay)²

[from Identity I, (x + y)² = x² + 2xy + y² and from identity II, (x – y)² = x² – 2xy + y²]

⇒ (ax + by)² – (ay – bx)²

⇒ ((ax + by) + (ay – bx))((ax + by) – (ay – bx))

We know, a³ – b³ = (a – b)(a² + ab + b²)

Given,

t⁹ – 512

= (t³)³ – (8)³

= (t³ – 8) (t⁶ + 8t³ + 64)

= [(t)³ – (2)³] (t⁶ + 8t³ + 64)

= (t – 2) (t² + 2t + 4) (t⁶ + 8t³ + 64)

We know, a² – b² = (a + b) (a – b)

Given,

729p⁶ – q⁶

= (27p³)² – (q³)²

= (27p³ + q³) (27p³ – q³)

Now, a³ – b³ = (a – b)(a² + ab + b²) and a³ + b³ = (a + b)(a² – ab + b²)

= [(3p)³ + (q)³] [(3p)³ – (q)³]

= (3p + q)(3p² – 3pq + q²) (3p – q)(3p² + 3pq + q²)

We know, a³ – b³ = (a – b)(a² + ab + b²)

Given,

8(p – 3)³ + 343

= [8(p – 3)]³ + (7)³

= [8(p – 3) + 7] [{8(p – 3)}² – 8(p – 3).7 + 7²]

= (8p – 24 + 7) [64(p – 3)² – 56(p – 3) + 49]

= (8p – 17) [64(p² – 6p + 9) – 56p + 168 + 49]

= (8p – 17) [64p² – 384p + 576 – 56p + 168 + 49]

= (8p – 17) [64p² – 440p + 793]

We know, a³ + b³ = (a + b)(a² – ab + b²)

Given,

Given,

(2a³ – b³)³ – b⁹

We know, (x – y)³ = x³ – y³ – 3xy(x – y)

Therefore,

= (2a³)³ – (b³)³ – 3(2a³)(b³)(2a³ – b³) – b⁹

= 8a⁹ – b⁹ – 6a³b³(2a³ – b³) – b⁹

= 8a⁹ – 12a⁶b³ + 6a³b⁶ – 2b⁹

= 2a⁹ – 2b⁹ + 6a⁹ – 12a⁶b³ + 6a³b⁶

= 2(a⁹ – b⁹) + 6a³(a⁶ – 2a³b³ + b⁶)

Now, we know: x³ – y³ = (x – y)(x² + xy + y²) and

(x – y)² = x² – 2xy + y²

= 2(a³ – b³)(a⁶ + a³b³ + b³) + 6a³(a³ – b³)²

= 2(a – b)(a² + ab + b²)(a⁶ + a³b³ + b⁶) + 6a³(a – b)²(a² + ab + b²)²

= 2(a – b)(a² + ab + b²)(a⁶ + a³b³ + b⁶ + 3a³(a – b)(a² + ab + b²))

= 2(a – b)(a² + ab + b²)( a⁶ + a³b³ + b⁶ + 3a³(a³ + a²b + ab² – a²b – ab² – b³)

= 2(a – b)(a² + ab + b²)( a⁶ + a³b³ + b⁶ + 3a³(a³ – b³) )

= 2(a – b)(a² + ab + b²)( a⁶ + a³b³ + b⁶ + 3a⁶ – 3a³b³ )

= 2(a – b)(a² + ab + b²)(4a⁶ – 2a³b³ + b⁶)

Given,

AR³ – Ar³ + AR²h – Ar²h

= A(R³ – r³ + R²h – r²h)

= A[(R³ – r³) + h(R² – r²)]

We know, a³ – b³ = (a – b)(a² + ab + b²) and a² – b² = (a + b)(a – b)

= A [(R – r) ( R² + Rr + r²) + h (R + r)(R – r)]

= A(R – r) [R² + Rr + r² + h (R + r)]

= A(R – r) [R(R + r) + r² + h(R + r)]

= A(R – r) (R + r) (R + r² + h)

Given,

a³ + 3a²b + 3ab² + b³ – 8

= (a + b)³ – (2)³

We know, a³ – b³ = (a – b)(a² + ab + b²)

= (a + b – 2) [(a + b)² + 2(a + b) + 2²]

We know, (a + b)² = a² + 2ab + b²

= (a + b – 2) [a² + 2ab + b² + 2a + 2b + 4]

32x⁴ – 500x

= 4x (8x³ – 125)

= 4x [(2x)³ – (5)³]

We know, a³ – b³ = (a – b)(a² + ab + b²)

= 4x (2x – 5) [(2x)² + 2x.5 + (5)²]

= 4x (2x – 5) (4x² + 10x + 25)

Given,

8a³ – b³ – 4ax + 2bx

= (2a)³ – (b)³ – 2x(2a – b)

We know, a³ – b³ = (a – b)(a² + ab + b²)

= (2a – b) ((2a)² + 2a.b + (b)²) – 2x(2a – b)

= (2a – b) (4a² + 2ab + b²) – 2x(2a – b)

= (2a – b) (4a² + 2ab + b² – 2x)

Given,

x³ – 6x² + 12x – 35

= x³ – 35 – 6x² + 12x

= x³ – 35 – 6x(x – 2)

x³ + y³ – 12xy + 64

⇒ x³ + y³ + 4³ – 12xy

⇒ x³ + y³ + 4³ - 3×4xy …Equation (i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

x³ + y³ + 4³ - 3×4xy = (x + y + 4) (x² + y² + 16 - xy - 4y - 4x)

(x + y + 4) (x² + y² + 16 - xy - 4y - 4x)

8x³ – y³ + 1 + 6xy

⇒ (2x) ³ + (- y) ³ + 1³ + 6xy

⇒ (2x) ³ + (- y) ³ + 1³ - 3×(2x)×(- y)×1 …Equation(i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ (2x) ³ + (- y) ³ + 1³ - 3×(2x)×(- y) = (2x – y + 1) ((2x)² + y² + 1² + 2xy + y - 2x)

⇒ (2x) ³ + (- y) ³ + 1³ - 3×(2x)×(- y) = (2x – y + 1) (4x² + y² + 1 + 2xy + y - 2x)

(2x – y + 1) (4x² + y² + 1 + 2xy + y - 2x)

8a³ – 27b³ – 1 – 18ab

⇒ (2a) ³ + (- 3b) ³ + (- 1)³ - 18ab

⇒ (2a) ³ + (- 3b) ³ + (- 1)³ - 3×(2a)×(- 3b)×(- 1) …Equation(i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ (2a) ³ + (- 3b) ³ + (- 1)³ - 3×(2a)×(- 3b)×(- 1) = (2a - 3b - + 1) (4a² + 9b² + 1 + 6ab - 3b - 2a)

(2a - 3b - 1) (4a² + 9b² + 1 - 6ab + 3b + 2a)

1 + 8x³ + 18xy – 27y³

⇒ 1³ + (2x) ³ + (- 3y) ³ + 18xy

⇒ 1³ + (2x) ³ + (- 3y) ³ + 3×(1)×(2x) (- 3y) …Equation(i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ 1³ + (2x) ³ + (- 3y) ³ + 3×(1)×(2x) (- 3y) = (1 + 2x - 3y) (1 + 4x² + 9y² - 2x + 6xy + 3y)

(1 + 2x - 3y) (1 + 4x² + 9y² - 2x + 6xy + 3y)

Let us consider (3a - 2b) = x, (2b - 5c) = y, (5c - 3a) = z

So x + y + z = 3a – 2b + 2b – 5c + 5c – 3a = 0

(3a – 2b) ³ + (2b – 5c) ³ + (5c – 3a) ³ = x³ + y³ + z³

Since x + y + z = 0, hence

x³ + y³ + z³ = 3xyz

⇒ (3a – 2b) ³ + (2b – 5c) ³ + (5c – 3a) ³ = 3×(3a – 2b)×(2b – 5c)× (5c – 3a)

3(3a – 2b) (2b – 5c) (5c – 3a)

Let us consider (2x – y) = a, (2y – x) = b

a + b = (2x – y) + (2y – x) = x + y

So we can say that

(2x – y) ³ – (x + y) ³ + (2y – x) ³ = a³ + b³ – (a + b) ³ …Equation (i)

Using the identity of (a + b) ³ we get

(a + b) ³ = a³ + b³ + 3ab(a + b)

⇒ - 3ab(a + b) = a³ + b³ - (a + b)³

Using the above identity in Equation (i)

(2x – y) ³ – (x + y) ³ + (2y – x) ³ = - 3ab(a + b)

⇒ (2x – y) ³ – (x + y) ³ + (2y – x) ³ = - 3×(2x – y)× (2y – x)× (x + y)

⇒ (2x – y) ³ – (x + y) ³ + (2y – x) ³ = 3×(2x – y)× (x - 2y)× (x + y)

3(2x – y) (x - 2y) (x + y)

a⁶ + 32a³ – 64

⇒ a⁶ + 8a³ – 64 + 24a³

⇒ (a²)³ + (2a) ³ + (- 4)³ - 3×(a²)×(2a)×(- 4) …Equation (i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ (a²)³ + (2a) ³ + (- 4)³ - 3×(a²)×(2a)×(- 4) = (a² + 2a - 4) (a⁴ + 4a² + 16 - 2a³ + 8a + 4a²)

⇒ (a² + 2a - 4) (a⁴ + 8a² + 16 - 2a³ + 8a)

(a² + 2a - 4) (a⁴ - 2a³ + 8a² + 8a + 16)

a⁶ – 18a³ + 125

⇒ a⁶ + 27a³ + 125 - 45a³

⇒ (a²)³ + (3a) ³ + 5³ - 45a³

⇒ (a²)³ + (3a) ³ + 5³ – 3× a²× 3a× 5 …Equation (i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ (a²)³ + (3a) ³ + 5³ – 3× a²× 3a× 5 = (a² + 3a + 5) (a⁴ + 9a² + 25 - 3a³ - 15a - 5a²)

⇒ (a²)³ + (3a) ³ + 5³ – 3× a²× 3a× 5 = (a² + 3a + 5) (a⁴ + 4a² + 25 - 3a³ - 15a)

(a² + 3a + 5) (a⁴ - 3a³ + 4a² - 15a + 25)

p³ (q – r) ³ + q³ (r – p) ³ + r³ (p – q) ³

⇒ (pq – pr) ³ + (qr – pq) ³ + (pr – qr) ³

Let us consider pq – pr = a, qr – pq = b, pr – qr = c

a + b + c = pq – pr + qr – pq + pr – qr = 0

Since a + b + c = 0, hence

a³ + b³ + c³ = 3abc

⇒ p³ (q – r) ³ + q³ (r – p) ³ + r³ (p – q) ³ = 3pqr(q - r) (r – p) (p – q)

3pqr (q - r) (r – p) (p – q)

…Equation (i)

We use the identity

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Using the above identity in Equation (i) we get

The given expression can be rewritten as:

(a + b)² – 5(a + b) + 6

Assume (a + b) = p,

⇒ p² – 5p+ 6

⇒ p² – 2p - 3p + 6

⇒ p(p - 2) -3(p - 2)

⇒ (p – 3)(p – 2)

On substituting the value of p, we get,

(a +b - 3)(a + b – 2)

The given expression can be rewritten as:

(x + 1)(3x – 1)(x + 2)(3x – 4) + 12

⇒ (3x² – x + 3x - 1)(3x² - 4x + 6x - 8) + 12

⇒ (3x² + 2x - 1)(3x² + 2x - 8) + 12

Let 3x² + 2x = p

⇒ (p - 1)( p – 8) + 12

⇒ p² – 9x + 8 + 12

⇒ p² – 9p + 20

⇒ p² – 5p – 4p + 20

⇒(p – 5)(p – 4)

On substituting the value of p, we get,

⇒ (3x² + 2x – 5)(3x² + 2x – 4)

⇒(3x² + 5x – 3x – 5)(3x² + 2x – 4)

⇒ (x(3x + 5) – 1(3x + 5))(3x² + 2x – 4)

⇒ (x – 1)(3x + 5) (3x² + 2x – 4)

As we know that,

a² – b² = (a – b)(a + b)

The given expression can be rewritten as:

x(x + 1)(x – 1)(x + 2) – 8

⇒ (x² + x)(x² + x – 2) – 8

Let x² + x = p

⇒ p(p – 2) – 8

⇒ p² – 2p – 8

⇒ p² – 4p + 2p – 8

⇒ p(p – 4) + 2(p – 4)

⇒ (p + 2)(p – 4)

On substituting the value of p, we get,

⇒ (x² + x + 2)(x² + x – 4)

As we know that,

x² – y² = (x – y)(x + y)

The given expression can be rewritten as:

7(a² + b²)² – 15(a² + b²)(a² – b²) + 8 (a² – b²)²

Let (a² + b²) = p and (a² – b²) = q

⇒ 7p² – 15pq + 8q²

⇒ 7p² – 7pq – 8pq + 8q²

⇒ 7p(p – q) – 8q(p – q)

⇒ (7p – 8q)(p – q)

On substituting the value of p and q, we get,

⇒ (7(a² + b²) - 8(a² – b²))( a² + b² – a² + b²)

⇒ (7a² + 7b² – 8a² + 8b²) (2b²)

⇒ 2b²(15b² – a²)

As we know that,

a² – b² = (a – b)(a + b)

The given expression can be rewritten as:

(x + 1)²(x – 1)² + 8x(x² + 1) + 19x²

Using (a – b)² = a² + b² – 2ab, and

(a + b)² = a² + b² + 2ab

⇒ (x² + 1 + 2x)(x² + 1 – 2x) + 8x(x² + 1) + 19x²

Let x² + 1 =p

⇒ (p + 2x)(p – 2x) + 8xp + 19x²

⇒ p² – 4x² + 8xp + 19x²

⇒ p² + 8xp + 15x²

⇒ p² + 3xp + 5xp + 15x²

⇒ p(p+3x) + 5x(p + 3x)

⇒ (p + 5x)(p + 3x)

On substituting the value of p, we get,

⇒ (x² + 5x + 1)(x² + 3x + 1)

The given expression can be rewritten as:

(a – 1)x² – x((a - 1) – (a - 2)) – (a – 2)

⇒ (a – 1)x(x – 1) + (a – 2)(x – 1)

⇒ (x – 1)(ax – x + a - 2)

Let p = (a – 1) and q = (a+1)

As we know that, a² – b² = (a – b)(a + b)

⇒ pq = a² + 1

The given expression can be rewritten as:

px² + pqxy + pq + qy²

⇒ px(x + qy) + y(x + qy)

⇒ (px + y)(x + qy)

On substituting the value of p and q, we get,

(ax – x + y)(x + ay + y)

The given expression can be rewritten as:

X² – qx – (p² – 5pq + 6q²)

⇒ x(x – q) – (p² – 2pq - 3pq + 6q²)

⇒ x² – qx – (p – 2q)(p – 3q)

⇒ x² + (p – 2q)x – (p – 3q)x – (p – 2q)(p – 3q)

⇒ (x – p + 2q)(x + p – 3q)

The given expression can be rewritten as:

As we know that, x² + y² - 2xy = (x – y)²

Let

⇒ 2p² – p – 3

⇒ 2p² – 3p + 2p – 3

⇒ p(2p – 3) + (2p – 3)

⇒ (p + 1) (2p – 3)

On substituting the value of p, we get,

The given expression can be rewritten as:

x(x – 1)y² + x²y – (x² – 1)y – x(x+1)

⇒ x² – xy² + x²y – x²y + y – x² – x

As we know that a² – b² = (a + b)(a – b)

⇒ xy[(x – 1)y + x] – (x + 1)[(x – 1)y + x]

⇒ (xy – x - 1)(xy + x – y)

Using the identity x² – y² = (x + y)(x – y)

⇒ (a + b)(a – b) = 11 × 9

Since 11 is a prime number therefore the factors can be equated,

⇒ a + b = 11

And a – b = 9

On solving above two equations we get,

a = 10 and b = 1

The given equation on LCM reduces to:

a² + b² = ab

Also we know that,

a³ + b³ = (a + b)(a² + b² – ab)

On substituting the value of a² + b² in above equation, we get,

⇒ (a + b) (ab – ab)

⇒ 0

The given expression can be rewritten as:

25³ – 75³ – (25 – 75)³ - 3(25)(75)(25 – 75)

As we know that,

(a – b)³ = a³ – b³ - 3ab(a - b)

⇒ a³ – b³ - (a – b)³ - 3ab(a - b) = 0

Since we can easily compare the above equation with the given expression by putting a = 25 and b = 75, the value of given expression is 0.

We know the identity that, if a + b+ c = 0, then,

a³ + b³ + c³ = 3abc

(x – 3) (x – a) = x² – (3 + a)x + 3a

⇒ 3a = 12

⇒ a = 4

Also, p = 3 + a

⇒ p = 3 + 4 = 7

Let p = (b² – c²), q = (c² – a²) and r = (a² – b²)

Since p + q+ r = 0 and we have an identity that if,

x + y + z = 0, the x³ + y³ + z ³ = 3xyz

⇒ p³ + q³ + r³ = 3pqr

Similarly in denominator of given fraction, we see that sum of all the individual terms is equal to 0, so same identity can be applied in denominator as well.

The fractions reduces to:

Also as, x² – y² = (x - y)(x + y)

As we know from the identity that,

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

Since a + b + c ≠ 0, that means,

a² + b² + c² – ab – bc – ca = 0

Multiplying both sides by 2, we get,

a² + b² – 2ab + b² + c² – 2bc + a² + c² – 2ac = 0

⇒ (a – b)² + (b – c)² + (a – c)² = 0

Now this is only possible if:

a – b = 0, b – c = 0 and a – c =0

⇒ a = b = c

224 can be written as 16 × 14

224 = (-16) × (-14)

Also using the identity, x² – y² = (x + y)(x – y)

⇒ a² – b² = (a + b)(a - b) = (-16) × (-14)

On equate the factors, as

a + b = -16 and (a – b) = -14 and solving these two equations, we get,

a = -15 and b = -1

As we know the identity that,

If l + m+ n = 0, then l³ + m³ + n³ – 3mnl = 0

Let p = (x - a), q = (x - b), r = (x – c)

p + q + r = (x – a + x – b + x – c) = 3x – a – b – c

also It is given 3x = a + b + c,

⇒ p + q + r = 0

⇒ p³ + q³ + r³ – 3pqr =0

On substituting the values of p, q and r in above equation, we get,

(x – a)³ + (x – b)³ + (x – c)³ – 3(x – a) (x – b) (x – c) = 0

∴ the required value is 0.

(2x– a) (x – 2) = 2x² – 4x – ax + 2a

⇒ 2x² –x(a + 4) + 2a

Since the above expression is identical to 2x² + px + 6, therefore their coefficients can be equated.

⇒ 2a = 6

⇒ a =3

Also, p = -(a + 4) = -7

∴ The values of a and p are 3 and -7 respectively.