
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 …
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No, Sir.
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