Let’s resolve into factors the following algebraic expressions:

i.

ii.

iii.

iv.

v.

vi.

vii.

viii.

ix.

x.

xi.

xii.

xiii.

xiv.

xv.

xvi.

xvii.

(i) 1000a³+27b⁶

= (10a)³ + (3b²)⁶

Using x³ + y³ = (x = y)(x² - xy + y²)

= (10a + 3b²)[(10a)² - (10a)(3b²) + (3b²)²]

= (10a + 3b²)(100a² - 30ab² + 9b⁴)

(ii) 1-216 z³

= 1 – (6z)³

Using x³ – y³ = (x – y)(x² + xy + y²)

= (1 – 6z)(1² + 1(6z) + (6z)²)

= (1 – 6z)(1 + 6z + 36z²)

(iii) m⁴ - m

= m(m³ – 1)

= m(m³ – 1³)

Using x³ – y³ = (x – y)(x² + xy + y²)

= m(m – 1)(m² + m + 1)

(iv) 192a³+3

= 3(64a³ + 1)

= 3[(4a)³ + 1³]

Using x³ + y³ = (x + y)(x² – xy + y²)

= 3(4a + 1)[(4a)² – 4a + 1]

= 3(4a + 1)(16a² – 4a + 1)

(v) 16a⁴x³ + 54ay³

= 2a(8a³x³ + 27y³)

= 2a[(2ax)³ + (3y)³]

Using x³ + y³ = (x + y)(x² – xy + y²)

= 2a(2ax + 3y)[(2ax)² – 2ax(3y) + (3y)²]

= 2a(2ax + 3y)(4a²x² – 6axy + 9y²)

(vi) 729a³ b³ c³-125

= (9abc)³ - 5³

= (9abc – 5)[(9abc)² + 9abc(5) + 5²]

= (9abc – 5)(81a²b²c² + 45abc + 25)

(vii)

Using x³ - y³ = (x - y)(x² + xy + y²)

(viii)

Using x³ – y³ = (x – y)(x² + xy + y²)

(ix)x³ + 3x²y + 3xy² + 2y³

= x³ + y³ + 3x²y + 3xy² + y³

= (x + y)³ + y³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (x + y + y)[(x + y)² - (x + y)y + y²]

[Using x³ – y³ = (x + y)(x² - xy + y²)]

= (x + 2y)(x² + y² - xy - y² + y²)

= (x + 2y)(x² – xy + y²)

(x) 1 + 9x + 27x² + 28x³

= 27x³ + 1 + 27x² + 9x + x³

= (3x)³ + 1³ + 3(3x)³(1) + 3(3x)(1)²

= (3x + 1)³ + x³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

Now, Using x³ + y³ = (x + y)(x² - xy + y²)

= (3x + 1 + x)[(3x + 1)² - (3x + 1)x + x²]

= (4x + 1)(9x² + 6x + 1 - 3x² - x + x²)

= (4x + 1)(7x² + 5x + 1)

(xi) x³ – 9y³ – 3xy(x-y)

x³ – y³ – 3xy (x – y) – 8y³

[∵ x³ – y³ = x³ – y³ – 3xy (x – y)]

= (x – y)³ – (2y)³

= (x – y – (2y)) [(x – y)² + (x – y) (2y) + (2y)²]

[Using x³ + y³ = (x + y) (x² – xy + y²)]

= (x -3 y) (x² + y² – 2xy + 2xy – 2y² + 4y²)

= (x - 3y) (x² + 3y²)

(xii)8 – a³ + 3a²b – 3ab² + b³

= b³ – a³ – 3ab² + 3a²b + 8

Now, As (x + y)³ = x³ + y³ + 3x²y + 3xy²

= (b – a)³ + 2³

Using x³ + y³ = (x + y) (x² - xy + y²)

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

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

(xiii) x⁶+3x⁴ b²+3x² b⁴+b⁶+a³b³

_{= (x}²)³ + (b²)³ + 3(x²)²(b²) + 3(x²)(b²)² + a³b³

_{As (x + y)}³ = x³ + y³ + 3x²y + 3xy²

_{= (x}² + b²)³ + (ab)³

_{Using x}³ + y³ = (x + y)(x² – xy + y²)

_{= (x}² + b² + ab)[(x² + b²)² - (x² + b²)ab + a²b²]

(xiv) x⁶ + 27

= (x²)³ + (3)³

Using x³ + y³ = (x + y)(x² – xy + y²)

= (x² + 3)[(x²)² - 3x² + 3²)

= (x² + 3)(x⁴ - 3x² + 9)

(xv)x⁶ – y⁶

= (x³)² – (y³)²

Using (a² – b²) = (a – b)(a + b)

= (x³ – y³)(x³ + y³)

Using x³ – y³ = (x – y)(x² + xy + y²) and

x³ + y³ = (x + y)(x² – xy + y²)

= (x – y)(x² + xy + y²)(x + y)(x² – xy + y²)

(xvi) x¹² – y¹²

= (x⁶)² – (y⁶)²

Using (a² – b²) = (a – b)(a + b)

= (x⁶ – y⁶)(x⁶ + y⁶)

= [(x³)² – (y³)^{2 �}][(x²)³ + (y²)³]

Using (a² – b²) = (a – b)(a + b)

= (x³ – y³)(x³ + y³)[(x²)³ + (y²)³]

Using x³ – y³ = (x – y)(x² + xy + y²) and

x³ + y³ = (x + y)(x² – xy + y²)

= (x – y)(x² + xy + y²)(x + y)(x² – xy + y²)(x² + y²)(x⁴ - x²y² + x⁶y⁶)

(xvii) m³ -n³-m(m²-n² )+n(m-n)²

Using x³ – y³ = (x – y)(x² + xy + y²) and

Using (a² – b²) = (a – b)(a + b)

= (m – n)(m² + mn + n²) – m(m – n)(m + n) + n(m – n)²

Taking (m – n) as common, we get

= (m – n)[m² + mn + n² – m(m + n) + n(m – n)]

= (m – n)(m² + mn + n² – m² – mn + mn – n²)

= (m – n)mn