Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Maths Solutions

Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to expand (a+b)^3 and (a-b)^3 easily. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Maths Solutions

Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Foundation Maths Solutions

Board ICSE
Publications Goyal brothers Prakshan
Subject Maths
Class 9th
Chapter-3 Expansions
Writer RS Aggrawal
Book Name Foundation
Topics More Identities
Academic Session 2024-2025

How to Expand (a+b)^3 and (a-b)^3 Easily

(a + b)3 Formula: The (a + b) whole cube formula or (a + b)3 formula represents the cube of a binomial. It represents the result of cubing the sum of ‘a’ and ‘b’.   (a + b)3 = a3 + 3a2b + 3ab2 + b3

(a – b)3 Formula:  The (a + b) whole cube formula or (a – b)3 formula represents the cube of a binomial. It represents the result of cubing the sum of ‘a’ and ‘b’.  (a – b)3 = a3 – 3a2b + 3ab2 – b3.

Exercise – 3B

Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Foundation Maths Solutions

Page- 52,53

Que-1: Expand :

(i) (3a+5b)³   (ii) (2p-3q)³   (iii) (2x+1/3x)³   (iv) (3ab-2c)³   (v) (3a-1/a)³  (vi) (1x/2 – 2y/3)³

Solution- (i) We know, (A+B)³ = A³+B³+3A²B+3AB²
Then,
(3a+5b)³
= (3a)³+(5b)³+3(3a)²5b+3(5b)²3a
= 27a³+125b³+3(9a²)(5b)+3(3a)(25b²)
= 27a³+125b³+135a²b+225ab².

(ii) (2p-3q)³
(a-b)³ = a³-b³-3a²b+3ab²
(2p-3q)³ = (2p)³-(3q)³-3(2p)²(3q)+3(2p)(3q)²
= 8p³-27q³-36p²q+54pq².

(iii) We know, (A+B)³ = A³+B³+3A²B+3AB²
Then,
(2x+1/3x)³
= (2x)³+(1/3x)³+3(2x)²(1/3x)+3(1/3x)²(2x)
= 8x³+1/27x³+3(4x²)(1/3x)+3(2x)(1/9x²)
= 8x³+1/27x³+4x+2/3x.

(iv) (3ab-2c)³
(a-b)³ = a³-b³-3a²b+3ab²
(3ab-2c)³ = (3ab)³-(2c)³-3(3ab)²(2c)+3(3ab)(2c)²
= 27a³b³-8c³-54a²b²c+36abc².

(v) (3a-1/a)³
(a-b)³ = a³-b³-3a²b+3ab²
(3a-1/a)³ = (3a)³-(1/a)³-3(3a)²(1/a)+3(3a)(1/a)²
= 27a³-1/a³-27a+9/a.

(vi) (1x/2-2y/3)³
(a-b)³ = a³-b³-3a²b+3ab²
(1x/2-2y/3)³ = (1x/2)³-(2y/3)³-3(1x/2)²(2y/3)+3(1x/2)(2y/3)²
= (1x³/8)+(8y³/27)-(1x²y/2)+(2xy²/3).

Que-2: If (4a+3b) = 10 and ab = 2, find the value of (64a³+27b³).

Solution-Given: (4a + 3b) = 10 and ab = 2
(64a3 + 27b3) = (4a)3 + (3b)3
As we know that x3 + y3 = (x + y) (x2 – xy + y2), so we have
(4a)3 + (3b)3
= (4a + 3b)(16a2 – 12ab + 9b2)
= 10(16a2 – 12×2 + 9b2)
= 10[(16a2 + 9b2) – 24]
= 10[(4a)2 + (3b)2] – 240
= 10[(4a + 3b)2 – 2(4a)(3b)] – 240
= 10[102 – 24×2] – 240
= 10(100 – 48) – 240
= 10 x 52 – 240
= 520 – 240
= 280 Ans.

Que-3: If (3x-2y) = 5 and xy = 6, find the value of (27x³-8y³).

Solution- Given : 3x-2y = 5  and  xy = 6
We know that,
(a+b)³ = (a-b)(a²+ab+b²)
Now,
27x³-8y³ = (3x)³ – (2y)³
where a = 3x  and b = 2y
(3x-2y){(3x)²+(2y)²+(3x.2y)}
[*(a-b)² = a²+b²-2ab
*(a-b)²+2ab = a²+b²-2ab+2ab
*(a-b)²+2ab= a²+b²]
= (3x-2y){(3x-2y)²+2×3x×2y+(6xy)}
= (3x-2y){(3x-2y)²+12xy+6xy}
= (3x-2y){(3x-2y)²+18xy}
put the values in equation :
= (5){(5)²+18×6}
= 5{25+108}
= 5{133}
= 665 Ans.

Que-4: If (a+3b) 6, show that a³+27b³+54ab = 216.

Solution- (a +3b) = 6
Cubing both sides we get.
(a +3b)³ = 6³
or, a³ + (3b)³ + 3.a.3b(a+3b) = 216
or, a³ + 27b³ + 9ab. 6 =216
or, a³ +27b³ + 54ab = 216 proved.

Que-5: If a+2b+3c = 0, show that a³+8b³+27c³ = 18abc.

Solution- a+2b+3c=0 then
a+2b=-3c
(a+2b)^3=(-3c)^3 (By cubing)
a^3+(2b)^3+3(a)(2b)(a+2b) = -27c^3
a^3+8b^3+6ab (-3c) = -27c^3 [a+2b=-3c]
a^3+8b^3-18abc = -27c^3
a^3+8b^3+27c^3 = 18abc
Hence proved..

Que-6: If (x+1/x) = 3, find the value of (x³+1/x³).

Solution- We know that,
a³+b³ = (a+b)³+3ab(a+b)
Here, a = x  and  b = 1x
Substitute these values in the above equation.
x³+1/x³ = (x+1/x)³+3(x+1/x)
= (3)³ + 3(3)
= 27+9
= 36 Ans.

Que-7: If (x-1/x) = 5, find the value of (x³-1/x³).

Solution- x−1/x = 5
Cubing both sides
(x−1/x)³ = (5)³
⇒ x³ − 1/x³ − 3(x−1/x) = 125
⇒ x³ − 1/x³ − 3×5 = 125
⇒ x³−1/x³ −15 = 125
⇒ x³−1/x³ = 125+15 = 140
∴ x³−1/x³ = 140 Ans.

Que-8: If (x-2/x) = 6, find the value of (x³-8/x³).

Solution- (x – 2/x) = 6.
As we know that,
Cubing on both sides of the equation, we get.
⇒ (x – 2/x)³ = (6)³.
⇒ (x)³ – 3(x)²(2/x) + 3(x)(2/x)² – (2/x)³ = (6)³.
⇒ x³ – (6x) + (12/x) – (8/x³) = 216.
⇒ x³ – 6(x – 2/x) – 8/x³ = 216.
Put the value of (x – 2/x) = 6 in the equation, we get.
⇒ x³ – 6(6) – 8/x³ = 216.
⇒ x³ – 36 – 8/x³ = 216.
⇒ x³ – 8/x³ = 216 + 36.
⇒ x³ – 8/x³ = 252 Ans.

Que-9: If (x+1/x) = 4, find the values of :  (i) (x³+1/x³)   (ii) (x-1/x)
(iii)
(x³-1/x³).

Solution- (i) Given, x+1/x = 4.
On taking cubes on both sides,
∴ (x+1/x)³ = 4³
∴ x³+1/x³+3.x.1/x(x+1x) = 64
∴ x³+1/x³ = 64−3.x.1/x(x+1x)
∴ x³+1/x³ = 64−3(x+1x)
= 64−3(4)
= 64−12 = 52 Ans.

(ii) Given x + 1/x = 4
On squaring both side, we get
(x + 1/x)² = 16
⇒ x² + 1/x² = 14 —- (i)
⇒ (x – 1/x)² = 14 – 2
⇒ (x – 1/x) = √12
⇒ x – 1/x = 2√3 Ans.

(iii) Given x + 1/x = 4
On squaring both side, we get
(x + 1/x)² = 16
⇒ x² + 1/x² = 14 —- (i)
⇒ (x – 1/x)² = 14 – 2
⇒ (x – 1/x) = √12
⇒ x – 1/x = 2√3 —- (ii)
On multiplying (i) and (ii), we get
x³ – 1/x³ – x + 1/x = 28√3
⇒ x³ – 1/x³ = 28√3 + x – 1/x
= x³ – 1/x³ = 28√3 + 2√3
Therefore x3 – 1/x= 30√3 Ans.

Que-10: If (a²+1/a²) = 23, find the values of : (i) (a+1/a)  (ii) (a³+1/a³)

Solution- (i) (a+1/a)² = a²+1/a²+2
⇒ (a+1/a)² = 23+2
⇒ (a+1/a)² = 25
⇒ a+1/a = √25
⇒ a+1/a = 5 Ans.

(ii) By using Binomial expansion, we get
a³+1/a³ = (a+1/a)(a²+1/a²-ab)
We know that a³+ b³= (a + b)(a²+b²-ab)
a³+1/a³ = 5(23 – 1)
a³+1/a³ = 5(22)
a³+1/a³ = 110 Ans.

Que-11: If (a-1/a) = √5, find the values of : (i) (a+1/a)  (ii) (a³+1/a³)

Solution- (i) (a-1/a) = √5
(a+1/a)^2 – (a -1/a)^2 = 4 (a*1/a)
(a+1/a)^2 – (√5)^2 = 4
(a+1/a)^2 – 5 = 4
(a+1/a)^2 = 5+4
(a+1/a)^2 = 9
(a+1/a) = √9
(a+1/a) = ±3 Ans.

(ii) (a+1/a) = 3
(a+1/a)³ = 3³
a³+1/a³+3(ax1/a)(a+1/a) = 27
a³+1/a³+3(1)(3) = 27
a³+1/a³+9 = 27
a³+1/a³ = 27 – 9
a³+1/a³ = 18 Ans.

Que-12: If (a²+1/a²) = 27, find the values of : (i) (a-1/a)  (ii) (a³-1/a³).

Solution- (i) a² + 1/a² = 27
Subtract 2 from both sides:
⇒ a² + 1/a² – 2 = 27 – 2
⇒ a² + (1/a)² – 2(a * 1/a) = 25
⇒ (a – 1/a)² = 25
⇒ a – 1/a = ±5 Ans.

(ii) Cube on both sides of a -1/a = 5:
⇒ (a – 1/a)³ = 5³
⇒ a³ – (1/a)³ – 3(a*1/a)(a – 1/a) = 125
⇒ a³ – 1/a³ – 3(1)(5) = 125
⇒ a³ – 1/a³ – 15 = 125
⇒ a³ – 1/a³ = 140 Ans.

Que-13: If (x²+1/25x²) = 8*(3/5), find the values of : (i) (x+1/5x) (ii) (x³+1/125x³)

Solution- (i) (x + 1/5x)² = x² + 1/25x² + 2/5
( x + 1/5x )² = 43/5 + 2/5
( x + 1/5x )² = (43 + 2)/5
( x + 1/5x )² = 45/5
( x + 1/5x )² = 9
x + 1/5x = √9
x + 1/5x = ±3 Ans.

(ii) When, x + 1/5x = 3,
( x + 1/5x )³ = ( 3 )³
x³ + 1/125x³ + 3/5( x + 1/5x ) = 27
x³ + 1/125x³ + (3/5) * 3 = 27
x³ + 1/125x³ = 27 – 9/5
x³ + 1/125x³ = ( 135 – 9 )/5
x³ + 1/125x³ = ±126/5 = ±25*(1/5) Ans.

Que-14: If (x+1/x)² = 3, show that (x³+1/x³) = 0.

Solution- (x+1/x)² = 3
x + 1/x = √3
(x+1/x)³ = x³ + (1/x)³ + 3(x)(1/x) (x + 1/x)
√3³ = x³ + 1/x³ + 3(√3)
x³ + 1/x³ = 3√3 – 3√3
x³ + 1/x³ = 0 Proved.

Que-15: If a/b = b/c, prove that (a+b+c)(a-b+c) = (a²+b²+c²)

Solution- As a, b, c, are in continued proportion
Let a/b = b/c = k
L.H.S. = (a + b + c) (a – b + c)
= (ck2 + ck + c)(ck2 – ck + c)
= c(k2 + k + 1)c(k2 – k + 1)
= c2(k2 + k + 1)(k2 – k + 1)
= c2(k4 + k2 + 1)
R.H.S. = a2 + b2 + c2
= (ck2)2 + (ck)2 + (c)2
= c2k4 + c2k2 + c2
= c2(k4 + k2 + 1)
∴ L.H.S. = R.H.S.

Que-16: Find the products using suitable identities :

(i) (3a+4b)(9a²-12ab+16b²)   (ii) (y-6/y)(y²+6+36/y²)

Solution- (i) (3a+4b)(9a²-12ab+16b²)
= (3a+4b)[(3a)² – (3a)(4b) + (4b)²]
= (3a)³ + (4b)³
= 27a³ + 64b³ Ans.

(ii) ( y – 6/y) (y²  + 6 + 36/y²)
36/y²  can be written as  (6/y)²
=  ( y – 6/y) (y²  + 6 + (6/y)²)
6 can be written as   y * 6/y
=  ( y – 6/y) (y²  + y*6/y + (6/y)²)
(a – b)(a² + ab + b²) = a³ – b³
a = y
b = 6/y
=  y³ – (6/y)³
= y³ – 216/y³ Ans.

Que-17: Simplify using suitable identity :

(3x-5y-4)(9x²+25y²+16+15xy-20y+12x)

Solution- (3x-5y-4) (9x²+25y²+15xy+12x-20y+16)
⇒ (3x-5y-4) (9x²+25y²+16+15xy-20y+12x)
⇒ {3x + (-5y) + (-4)} {(3x)² + (-5y)² + (-4)² – (3x) (-5y) – (-5y) (-4) – (-4) (3x)}
⇒ (3x)³ + (-5y)³ + (-4)³ – 3 (3x) (-5y) (-4)
[As we know a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² -ab – bc – ca)]
= 27x³ – 125y³ – 64 – 180xy Ans.

– : End of Expansions Class 9 RS Aggarwal Exe-3B Goyal Brothers ICSE Maths Solutions : –

Return to :- RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)

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