**Find the following products:**

**(i)(x + 4) (x + 5) (x + 2) **

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

a = 4, b = 5 and c =2

(x + 4) (x + 5) (x + 2) = x^{3} + x^{2}(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2

= x^{3} + 11x^{2} + x (20 + 10 + 8) +40

= x^{3} + 11x^{2} + 38x +40

(ii) (y + 3) (y + 2) ( y – 1)

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc we get

x = y, a = 3, b = 2 and c = -1

(y + 3) (y + 2) (y – 1) = y^{3} + y^{2} (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)

= y^{3} + 4y^{2} + y (6 – 2 – 3) – 6

= y^{3} + 4y^{2} + y – 6

**(iii) (a + 2) (a – 3) (a + 4) **

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = a, a = 2, b = –3 and c = 4

(a + 2) (a – 3) (a – 4) = a^{3} + a^{2} (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4

= a^{3} + 3a^{2} + a (–6 – 12 + 8) – 24

= a^{3} + 3a^{2} – 10a – 24

**(iv) (m – 1) (m – 2) (m – 3) **

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = m, a = –1, b = –2 and c = –3

(m – 1) (m – 2) (m – 3) = m^{3 }+ m^{2} (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +

(– 3)(–1)] + (–1) (–2) (–3)

= m^{3} + m^{2} (–6) + m [2 + 6 + 3] – 6

= m^{3} – 6m^{2} + 11m – 6

**(v) ( √****𝟐**** + √𝟑****) (√****𝟐****+ √****𝟓****) (√****𝟐****+ √****𝟕**** ) **

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = 2, a = 3, b = 5 and c = 7

(√2 + √3) (√2+ √5) (√2+ √7 ) = (√2)^{3} + (√2)^{2} [√3 + √5+ √7]+ √2 [√3. √5 + √5. √7 + √7. √3] + √3. √5. √7

= 2√2 + 2(√3 + √5 + √7) + √2 (15+ √35 + √21) + √105

**(vi) 105 x 101 x 102 **

We can write this as

(100 + 5) (100 + 1) (100 + 2)

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = 5, b = 1 and c = 2

(100 + 5) (100 + 1) (100 + 2) = 100^{3} + 100^{2} (5 + 1 + 2) + 100 (5.1 + 1.2

+ 2.5) + 5.1.2

= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10

= 1000000 + 80000 + 1700 +10

= 1081710

**(vii) 95 x 98 x 103 **

We can write this as

(100 – 5) (100 – 2) (100 + 3)

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = -5, b = -2 and c = 3

(100 – 5) (100 – 2) (100 + 3) = 100^{3} + 100^{2} (–5 – 2 +3) + 100(–5) (–2) + (–2) 3

+ 3 (-5) + (-5) (-2) 3

= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30

= 1000000 – 40000 – 1100 +30

= 958930

**(viii) 1.01 x 1.02 x 1.03 **

We can write this as

(1 + 0.01) (1 + 0.02) (1 + 0.03)

Using (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc

we get

x = 1, a = 0.01, b = 0.02 and c = 0.03

(1 + 0.01) (1 + 0.02) (1 + 0.03) = 1^{3} + 1^{2} (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +

(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)

= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006

= 1.061106

**Find the coefficients of x2 and x in the following:**

**(i) (x + 4) (x + 1) (x + 2) **

= x^{3} + x^{2} (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2

= x^{3} + 7x^{2} + 14x + 8

Coefficient of x^{2} is 7

Coefficient of x is 14

**(ii) (x – 5) (x – 6) (x – 1) **

= x^{3} + x^{2} (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)

= x^{3}– 12x^{2} + x (30 + 6 + 5) – 30

= x^{3}– 12x^{2} + 41x – 30

Coefficient of x^{2} is –12 and x is 41

**(iii) (2x + 1) (2x – 2) (2x – 5) **

= (2x)^{3} + (2x)^{2} [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)

= 8x^{3} + 4x^{2} (–6) + 2x [–2 +10–5] + 10

= 8x^{3} – 24x^{2} + 6x + 10

Coefficient of x^{2} is –24 and x is 6

**(iv) ( **^{𝐱}**/ _{𝟐}**

**+ 1) (**

^{𝐱}**/**

_{𝟐}**+ 2) (**

^{𝐱}**/**

_{𝟐}**+ 3)**

= (^{𝐱}/ )3 + (^{𝐱}/_{𝟐})2 [1 + 2 + 3] + ^{𝐱}/_{𝟐} [1.2 + 2.3 + 3.1] + 1.2.3

= x^{3}/_{8} + (x^{2}/_{4} )(6) + ^{𝐱}/_{𝟐} (2 + 6 + 3) + 6

= x^{3}/_{8} + ^{3}/_{2} x^{2} + ^{11}/_{2} x + 6

Coefficient of x^{2} is 𝟑𝟐 and x is 𝟏𝟏𝟐

**The length, breadth and height of a cuboids are (x +3), (x – 2) and (x -1) respectively. Find its volume.**

**Solution:**

Volume of a cuboids = length x breadth x height

V = (x +3) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc we get

V = x^{3} + x^{2} (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)

= x^{3} – 0x^{2} + X (–6 + 2 – 3) + 6

= x^{3} – 7x^{2} + 6

**The length, breadth and height of a metal box are cuboid are (x +5), (x – 2) and (x – 1) respectively. What is its volume?**

Solution:

Volume of the metal box = length x breadth x height

V = (x +5) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc we get

V = x^{3} + x^{2} + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)

= x^{3} + 2x^{2} + x [–10 + 2 – 5] + 10

= x^{3} + 2x^{2} – 13x + 10

**Prove that**

**(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc **

**[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d] **

Solution:

x^{3} + x^{2} (a + b + c) + x (ab + bc + ca) + abc we get

- H.S. = (a + b + c)
^{3}+ (a + b + c)^{2}(–c – a – b) + (a + b + c) [(–c) (–a) +

(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)

= (a + b + c)^{3}-(a + b + c)^{2}[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc

= (a + b + c)^{3}– (a + b + c)^{3} + (a + b + c) (ac + ab + bc) – abc

= (a + b + c) (ac + ab + bc) – abc

= R. H. S.

**Find the cubes of the following:**

**(i) (2x +y) ^{3} **

Solution:

Using (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} we get

a = 2x, b = y

(2x +y)^{3} = (2x)^{3} + 3(2x)^{2} y + 3 (2x) y^{2} +y^{3}

= 8x^{3} + 12 x^{2}y + 6 xy^{2} +y^{3}

**(ii) (2x + 3y) ^{3} **

Solution:

Using (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} we get

a = 2x, b = 3y

(2x + 3y)^{3} = (2x)^{3} + 3(2x)^{2} (3y) + 3 (2x) (3y)^{2} + (3y)^{3}

= 8x^{3} + 36 x^{2}y + 54 xy^{2 }+ 27y^{3}

**(viii) 101 ^{3} **

Solution:

We write 101 as (100 + 1)^{3}

Using identity (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} we get

a = 100, b = 1

(100 + 1)^{3} = 100^{3} + 3. 100^{2} + 3. 100.1^{2} + 1^{3 }

= 1000000 + 30000 + 300 + 1

= 1030301

**(viii) 2.1 ^{3} **

Solution:

We write 2.1 (2 + 0.1)^{3}

Using identity (a + b)^{3} = a^{3 }+ 3a^{2}b + 3ab^{2} + b^{3 }we get

a = 2, b = 0.1

(2 + 0.1)^{3} = 2^{3} + 3 x 2^{2}(0.1) + 3 x 2 x (0.1)^{2} + (0.1)^{3 }

= 8 + 1.2 + 0.06 + 0.001

= 9.261

**Find the cubes of the following:**

**(i) (2a – 3b) ^{3}**

Solution:

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = 2a, b = 3b

(2a – 3b)^{3} = (2a)^{3} – 3 (2a)^{2}(3b) + 3 (2a)(3b)^{2} – (3b)^{3}

= 8a^{3} – 36a^{2}b + 54ab^{2} -27b³

**(ii) ( x – **^{𝟏}**/ _{𝐱}**

**)**

^{3}Solution:

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = x, b = ^{𝟏}/_{𝐱}

(x – ^{𝟏}/ )^{3} = x^{3} – 3x^{2} ^{𝟏}/_{𝐱} + 3x(^{𝟏}/_{𝐱})^{3} – (^{𝟏}/_{𝐱})^{3 }= x^{3} – 3x + ^{3x}/ x^{2} – ^{1}/x^{3}

= x^{3} – 3x + ^{3}/_{x} – ^{1}/x^{3}

**(iii) (√3 x – 2) ^{2} **

Solution:

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = √3 x, b = 2

(√3 x – 2)^{2} = (√3x)^{3} – 3 (√3x)^{2} .2 + 3. √3x x 2^{2} – 2^{3}

= 3√3 x^{3 }– 6. 3 x^{2} + 12√3 x – 8

=3√3 x^{3} – 18x^{2} + 12√3 x – 8

**(iv) (2x – √5) ^{3} **

Solution:

Using (a – b)^{3} = a^{3 }– 3a^{2}b + 3ab^{2} – b^{3} we get

a = 2x, b = √5

(2x – √5)^{3} = (2x)^{3} – 3(2x)^{2} √5 + 3. 2x. √5)^{2} – (√5)^{3}

= 8x^{3} – 12√5x^{2} + 30x – 5√5

**(v) 49 ^{3 }**

Solution:

We can write 49 = 50 – 1

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = 50, b = 1

(50 – 1)^{3} = 50^{3} – 3.50^{2}.1 + 3.50.1^{2 }– 1^{3 }

= 125000 – 3 x 2500 + 150 – 1

= 125000 – 7500 +149

= 117649

**(vi) 18 ^{3} **

Solution:

Let us write 18 = 20 – 2

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = 20, b = 2

(20 – 2)^{3} = 20^{3} – 3.20^{2}.2 + 3.20.2^{2} – 2^{3}

= 8000 – 6×400 + 60×4 – 8

= 8000 – 2400 +240 – 8

= 5832

**(vii) 95 ^{3} **

Solution:

We write 95 = 100 – 5

Using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = 100, b = 5

(100 – 5)^{3} = 100^{3} – 3.100^{2}.5 + 3.100.5^{2} – 5^{3}

= 1000000 – 150000 + 7500 – 125

= 857375

**(viii) 108 ^{3} **

Solution:

We write 108^{3} = (110 – 2)

Using (a – b)^{3 }= a^{3} – 3a^{2}b + 3ab^{2} – b^{3} we get

a = 110, b = -2

(110 – 2)^{3} = 110^{3} – 3. (110)^{2}.2 + 3.110×2^{2} – 2^{3}

= 1331000 – 72600 + 1320 -8

= 1259712

**If x +**^{𝟏}**/**_{𝐱}**= 3, prove that x**^{3}+ 𝟏**/𝐱**^{𝟑}**= 18.**

Solution:

Given x + ^{1}/_{x} = 3

Cubing both sides we get

(x + ^{1}/_{x} )^{3} = 3^{3}

Using (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} we get

a = x b = 1/x

(x + 1/x )^{3} = (x)^{3} + ( 1/x )^{3} + 3x. ^{1}/_{x} (x + ^{1}/_{x} )

27 = x^{3} + ^{1}/x^{3} + 3 (3)

x^{3} + 1/x^{3} = 27 – 9

= x^{3} + ^{1}/x^{3} = 18

**If p + q = 5 and pq = 6, find p**^{3}+ q^{3}

Solution:

(p + q)^{3} = p^{3} + 3pq (p + q) + q^{3}

5^{3} = p^{3} + 3.6 (5) + q^{3}

125 = p^{3} + 90 + q^{3}

p^{3} + q^{3} = 125 – 90

p^{3} + q^{3} = 35

**If a – b = 3 and ab = 10, find a**^{3}– b^{3}

Solution:

Given a – b = 3 and ab = 10

(a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)

3^{3} = a^{3} – b^{3} – 3.10 (3)

27 = a^{3} – b^{3} – 90

a^{3} – b^{3} = 27 + 90

a^{3} – b^{3} = 117

**If a**^{2}+^{𝟏}**/𝐚**^{𝟐}**= 20 and a**^{3}+^{𝟏}**/𝐚**^{𝟑}**= 30, find a + 𝟏****/𝐚**

Solution:

a^{3} + 1/a^{3} = (a + ^{1}/_{a}) (a^{2} + ^{1}/a^{2} – a x ^{1}/_{a} )

30 = (a + ^{1}/_{a} ) = (20 – 1)

30 = (a +^{1}/_{a} ) x 19

^{30}/_{19} = a + ^{1}/_{a}

a + ^{1}/_{a} = ^{𝟑𝟎}/_{𝟏𝟗}

is the poly mult a Vedic method? if so, what’s the source? I have Vedic Math,1992, and don’t recall seeing that method in that book …

No, Sir.