Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.3

Free PDF download of Balaji Publications Mathematics Class 9 Solutions Chapter 1 Real Numbers Ex 4.3 for all the questions enlisted under the chapter. All questions are solved using step by step approach, solving the questions from the Class 9 Maths for helps in grasping the concepts correctly.

Balaji Class 9 Maths Solutions Chapter 4 Algebraic Identities Ex 4.3 बीजगणितीय सर्वसमिकाऐं

Ex 4.3 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
(2x + 1)³ का मान ज्ञात कीजिए। [NCERT]
हलः
(2x + 1)³ = (2x)³ + (1)³ + 3 × 2x × 1(2x + 1)
= 8x³ + 1 + 12x² + 6x

प्रश्न 2.
(2a – 3b) का मान ज्ञात कीजिए। [NCERT]
हलः
(2a – 3b)³ = (2a)³ + (-3b)³ + 3(2a) (-3b)(2a – 3b)
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²

Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न – I (Short Answer Type Questions – I)

प्रश्न 3.
निम्न व्यंजकों के घन का मान ज्ञात कीजिए। (NCERT Exemplar)
(i) $\left(\frac{1}{x}+\frac{y}{3}\right)$
(ii) (4 – $\frac{1}{3 x}$ )
(iii) (3x – 2y)
हलः

प्रश्न 4.
निम्न को सरल कीजिए।
(i) (4x + 2y)³ + (4x – 2y)³
(ii) (4x + 2y)³ – (4x – 2y)³
(iii) (a + 3)³ + (a – 3)³
हलः
(i) (4x + 2y)³ = (4x)³ + (2y)³ + 3. (4x) . (2y) (4x + 2y)
= 64x³ + 8y³ + 24xy (4x + 2y)
= 64x³ + 8y³ + 96x²y + 48xy²

(4x – 2y)³ = (4x)³ + (-2y)³ + 3. (4x) (-2y) (4x -2y)
= 64x³ – 8y³ – 24xy (4x -2y) –
= 64x³ – 8y³ – 96x²y + 48xy²
इसलिए (4x + 2y)³ + (4x – 2y)³ = 128x³ +96xy²

(ii) (4x + 2y)³ – (4x – 2y)³
= 64x³ + 8y³ + 96x²y + 98xy² – (64x³ – 8y³ – 96x²y + 48xy²)
= 64x³ + 8y³ + 96x³²y + 48xy² – 64x³ + 8y³ + 96x²y – 48xy²
= 16y³ + 192x²y

(iii) (a + 3)³ = a³ + 27 + 3a . 3(a + 3)
= a³ + 27 + 9a² + 27a
(a – 3)³ = a³ – 27 – 3a – 3(a – 3)
= a³ – 27 – 9a² + 27a
∴ (a + 3)³ + (a – 3)³ = 2a³ +54a

Ex 4.3 Algebraic Identities लघु उत्तरीय प्रश्न – II (Short Answer Type Questions – II)

प्रश्न 5.
यदि a – $\frac{1}{a}$ = 7, तब $a^{3}-\frac{1}{a^{3}}$ का मान ज्ञात कीजिए।
हलः

प्रश्न 6.
यदि a – b = 4 व ab = 21, तब a³ – b³ का मान ज्ञात कीजिए।
हलः
a – b = 4
वर्ग करने पर
a² + b² – 2ab = 16
a² + b² – 2 × 21 = 16
a² + b² = 16 + 42 = 58
अब a³ – b³ = (a – b)(a + b² + ab)
= (4)(58 + 21)
= (4)(79) = 316

प्रश्न 7.
यदि x + y = 10 व xy = 21, तब x³ + y³ का मान ज्ञात कीजिए।
हलः
x + y = 10 ……………. (1)
घन करने पर
(x + y)³ = (10)³
x³ + y³ + 3xy (x + y) = 1000
x³ + y³ + 3. 21 (10) = 1000
x³ + y³ + 630 = 1000
x³ + y ³ = 1000 – 630 = 370

प्रश्न 8.
यदि 3x – 2y =11 व xy = 12, तब 27x³ – 8y³ का मान ज्ञात कीजिए।
हल:
∵ 3x – 2y = 11 …………….. (1)
वर्ग करने पर
9x² + 4y² – 12xy = 121
9x² + 4y² -12 × 12 = 121
9x² + 4y² = 121 + 144 = 265
27x³ – 8y³ = (3x)³ – (2y)³
= (3x – 2y)(9x² + 4y² + 6xy)
= (11)(265 + 6 × 12)
= (11)(265 + 72)
= (11) (337) = 3707

प्रश्न 9.
यदि $a^{2}+\frac{1}{a^{2}}$ = 98, तब $a^{3}+\frac{1}{a^{3}}$ का मान ज्ञात कीजिए।
हल:

प्रश्न 10.
यदि $a^{2}+\frac{1}{a^{2}}$ = 51, तब $a^{3}-\frac{1}{a^{3}}$ का मान ज्ञात कीजिए। .
हल:

Ex 4.3 Algebraic Identities दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

प्रश्न 11.
निम्न के मान ज्ञात कीजिए।
(i) (99)³
(ii) (9.9)³
(iii) (10.4)³
(iv) (598)³
(v) (402)³
(vi) (1002)³
हलः
(i) (99)³ = (100 – 1)³
___ = (100) – (1)³ – 3 × 100 × 1(100 – 1)
= 1000000 – 1 – 30000+ 300 = 970299

(ii) (9.9)³ = (10 – 0.1)³
= (10)³ – (0.1)³ – 3 × 10 × 0.1(10 – 0.1)
= 1000 – 0.001 – 3(9.9)
= 1000 – 0.001 – 29.7 = 970.299

(iii) (10.4)³ = (10 + 0.4)³
= (10)³ + (0.4)³ + 3 × 10 × 0.4(10 + 0.4)
= 1000+ 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 =1124.864

(iv) (598)³ = (600-2)³
= (600)³ – (2)³ – 3 × 600 × 2(600 – 2)
= 216000000 – 8-21,60,000 + 7200
=21,38,47,192

(v) (402)³ = (400 + 2)³
= (400)³ + (2)³ + 3 × 400 × 2(400 + 2)
= 64000000 + 8 + 960000 + 4800
= 64964808

(vi) (1002)³ = (1000 + 2)³
= (1000)³ + (2)³ + 3 × 1000 × 2(1000 + 2)
= 1000000000 + 8+ 6000000 + 12000
= 1006012008

प्रश्न 12.
निम्न के मान ज्ञात कीजिए।
(i) (46)³ (34)3
(ii) (23)³ – (17)³
(iii) (111)³ – (89)³
हल:
(i) (46)³ + (34)³
= (40 + 6)³ + (40-6)³
= (40)³ + (6)³ + 3 × 40 × 6(40 + 6) + (40)³ – (6)³ – 3 × 40 × 6(40 – 6)
= 2(40)³ + 3 × 40 × 6[46 – 34]
= 2(40)³ + 720 × 12
= 128000 + 8640 = 136640

(ii) (23)³ – (17)³
= (20 + 3)³ – (20 – 3)³
= (20)³ + (3)³ + 3 × 20 × 3(20 + 3) – (20)³ + (3)³ + 3 × 20 × 3(20 – 3)
= 2(3)³ + 3 × 20 × 3(23 + 17)
= 2 × 27 + 180(40) = 54+ 7200 = 7254

(iii) (111)³ – (89)³
= (100 + 11)³ – (100 – 11)³
= (100)³ + (11)³ + 3 × 100 × 11 (100 + 11) – (100)³ + (11)³ + 3 × 100 × 11(100 – 11)
= 2(11)³ + 3 × 100 × 11(111 + 89)
= 2(1331) + 3 × 100 × 11 × 200
= 2662 + 660000 = 662662

प्रश्न 13.
यदि 3x +2y = 20 व xy = $\frac{14}{9}$ तब 27x³ + 8y³ का मान ज्ञात कीजिए।
हलः
3x + 2y = 20 …………… (1)
वर्ग करने पर
9x² + 4y² + 12xy = 400
9x² +4y² = 400 – 12xy
27x³ + 8y³
= (3x)³ + (2y)³
= (3x + 2y) (9x² + 4y² – 6xy)
= (3x + 2y)(400 – 12xy – 6xy)
= (3x + 2y)(400 – 18xy) = (20)( 400 – 18 × $\frac{14}{9}$ )
= (20) (400 – 28) = (20) (372) = 7440