8th Maths : Chapter 3 : Algebra : Cubic Identities : Exercise 3.3: Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.3**

**1. Expand**

**(i) (3 m **

**(ii) (5 p **

**(iii) (2 n **

**(iv) 4 p ^{2} **

**Solution:**

**(i)**** (3 m + 5)^{2}**

Comparing (3*m* + 5)^{2} with (*a + b*)^{2}
we have *a* = 3*m* and *b* = 5

(*a + b*)^{2} = *a*^{2} + 2*ab* +
*b*^{2}

(3*m* + 5)^{2} = (3*m*)^{2} + 2 (3*m*)
(5) + 5^{2}

= 3^{2} *m*^{2} + 30 *m* + 25 = 9*m*^{2}
+ 30*m* + 25

**(ii)**** (5 p − 1)^{2}**

Comparing (5*p* − l)^{2} with (*a − b*)^{2}
we have *a* = 5*p* and *b* = 1

(*a − b*)^{2} = *a*^{2} − 2*ab* +
*b*^{2}

(5*p* − l)^{2} = (5*p*)^{2} − 2 (5*p*)
(l) + l^{2}

= 5^{2}*p*^{2} − 10*p* + 1 = 25*p*^{2}
− 10*p* + 1

**(iii)**** (2 n − 1) (2n + 3)**

Comparing (2*n* − 1) (2*n* + 3) with (*x* + a) (*x + b*) we have *a* =
− l; *b* = 3

(*x + a*) (*x* + *b*) = *x*^{2} + (*a + b*)*x*
+ *ab*

(2*n* + (− 1)) (2*n* + 3) = (2*n*)^{2} +
(−1 + 3) 2*n* + (−1) (3)

= 2^{2}*n*^{2} + 2 (2*n*) − 3= 4*n*^{2}
+ 4*n* − 3

**(iv)**** 4 p^{2} − 25q^{2} = (2p)^{2}
− (5q)^{2}**

Comparing (2*p*)^{2} − (5*q*)^{2} with
*a*^{2} − *b*^{2} we have *a* = 2*p* and *b*
= 5*q*

(*a*^{2} − *b*^{2}) = (*a + b*) (*a
− b*) = (2*p* + 5*q*) (2*p* − 5*q*)

** **

**2. Expand**

**(i) (3 ****+**** m)^{3}**

**(ii) (2 a **

**(iii) (3 p **

**(iv) (52) ^{3}**

**(v) (104) ^{3}**

**Solution:**

**(i)**** (3 + m)^{3}**

Comparing (3 + *m*)^{3 }with (*a + b*)^{3}
we have* a* = 3 ; *b* = *m*

(*a + b*)^{3} = *a*^{2} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{3}

(3* + m*)^{3} = 3^{3} + 3(3)^{2}(*m*)
+ 3(3) *m*^{2} + *m*^{3}

= 27 + 27*m* + 9*m*^{2} + *m*^{3 }=
*m*^{3 }+ 9*m*^{2} + 27*m* + 27

**(ii) ****(2 a + 5)^{3}**

Comparing (2*a* + 5)^{3} with (*a + b*)^{3}
we have *a* = 2*a*, *b* = 5

(*a + b*)^{3} = *a*^{3} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{3}

= (2*a*)^{3} + 3(2*a*)^{2} 5 + 3 (2*a*)
5^{2} + 5^{3}

= 2^{3}*a*^{3} + 3(2^{2}*a*^{2})
5 + 6*a* (25) + 125

= 8* a*^{3 }+ 60* a*^{2} + 150*a*
+ 125

**(iii) ****(3 p + 4q)^{3}**

Comparing (3*p* + 4*q*)^{3 }with (*a + b*)^{3}
we have *a* = 3*p* and *b* = 4*q*

(*a + b*) = *a*^{3} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{2}

(3*p* + 4*q*)^{3
}= (3*p*)^{3}+ 3(3*p*)^{2 }(4*q*) +
3(3*p*) (4*q*)^{2} + (4*q*)^{3}

= 3^{3}*p*^{3 }+ 3 (9*p*^{2})
(4*q*) + 9*p* (16*q*^{2}) + 4^{3}*q*^{3}

= 27*p*^{3 }+ 108*p*^{2}*q* + 144*pq*^{2}
+ 64*q*^{3}

**(iv)**** (52) ^{3} = (50+ 2)^{3}**

Comparing (50 + 2)^{3} with (a + b)^{3} we have *a*
= 50 and *b* = 2

(*a + b*)^{3} = *a*^{3} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{3}

(50 + 2)^{3} = 50^{3} + 3 (50)^{2} 2 + 3
(50)(2)^{2} + 2^{3}

52^{3} = 125000 + 6(2,500) + 150(4) + 8

= 1,25,000 + 15,000 + 600 + 8

52^{3} = 1,40,608

**(v)**** (104) ^{3} = (100+ 4)^{3}**

Comparing (100 + 4)^{3} with (*a + b*)^{3}
we have *a *= 100 and *b* = 4

(*a + b*)^{3} = *a*^{3} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{3}

(100 + 4)^{3} = (100)^{3} + 3 (100)^{2}
(4) + 3 (100) (4)^{2} + 4^{3}

= 10,00,000 + 3 (10000) 4 + 300 (16) + 64

= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

** **

**3. Expand**

**(i) (5 ****−**** x)^{3}**

**(ii) (2 x **

**(iii) ( ab **

**(iv) (48) ^{3}**

**(v) (97 xy)^{3}**

**Solution:**

**(i)**** (5 − x)^{3} **

Comparing (5 − *x*)^{3}** **with (*a − b*)^{3}
we have *a* = 5 and *b* = *x*

(*a − b*)^{3} = *a*^{3} − 3*a*^{2}*b*
+ 3*ab*^{2} − *b*^{3}

(5 − *x*)^{3}** **= 5^{3} − 3 (5)^{2}
(*x*) + 3(5)(*x*^{2}) − *x*^{3}

= 125 − 3(25)(*x*) + 15*x*^{2} − *x*^{3}
= 125 − 75*x* + 15 *x*^{2} − *x*^{3}

**(ii) ****(2 x − 4y)^{3} **

Comparing (2*x* − 4*y*)^{3} with (*a − b*)^{3}
we have *a* = 2*x* and *b* = 4*y*

(*a − b*)^{3} = *a*^{3} − 3*a*^{2}*b*
+ 3*ab*^{2} − *b*^{3}

(2*x* − 4*y*)^{3} = (2*x*)^{3} − 3(2*x*)^{2}
(4*y*) + 3(2*x*) (4*y*)^{2} − (4*y*)^{3}

= 2^{3} *x*^{3} − 3(2^{2} *x*^{2})
(4*y*) + 3(2*x*) (4^{2} *y*^{2}) − (4^{3}*y*^{3})

= 8*x*^{3} − 48*x*^{2}*y* + 96*xy*^{2}
− 64*y*^{3}

**(iii) ****( ab − c)^{3}**

Comparing (*ab − c*)^{3 }with (*a − b*)^{3}
we have *a* = *ab *and *b* = *c*

(*a − b*)^{3} = *a*^{3} − 3*a*^{2}*b*
+ 3*ab*^{2} − *b*^{3}

(*ab − c*)^{3 }= (*ab*)^{3} − 3(*ab*)^{2}
*c* + 3*ab* (*c*)^{2} − *c*^{3}

= *a*^{3}*b*^{3} − 3(*a*^{2}*b*^{2})
*c* + 3*abc*^{2} − *c*^{3}

= *a*^{3}*b*^{3} − 3*a*^{2}*b*^{2}
*c* + 3*abc*^{2} − *c*^{3}

**(iv)**** (48) ^{3} = (50 −2)^{3}**

Comparing (50 − 2)^{3} with (*a − b*)^{3}
we have *a* = 50 and *b* = 2

(*a − b*)^{3} = *a*^{3} − 3*a*^{2}*b*
+ 3*ab*^{2} − *b*^{3}

(50 − 2)^{3} = (50)^{3} − 3(50)^{2} (2)
+ 3 (50)(2)^{2} − 2^{3}

= 1,25,000 − 15000 + 600 − 8 = 1,10,000 + 592

= 1,10,592

**(v) ****(97 xy)^{3}**

(97*xy*)^{3 }= 97^{3} *x*^{3}*y*^{3}
= (100 − 3)^{3} *x*^{3}*y*^{3} ... (1)

Comparing (100 − 3)^{3} with (*a − b*)^{3}
we have *a* = 100, *b* = 3

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

(100 − 3)^{3} = (100)^{3} − 3(100)^{2}
(3) + 3 (100)(3)^{2} − 3^{3}

97^{3} = 10,00,000 − 90000 + 2700 − 27

97^{3} = 910000 + 2673

97^{3} = 912673

97*x*^{3}*y*^{3} = 912673*x*^{3}*y*^{3}

** **

**4. Simplify ( p **

**Solution:**

(*p* − 2)(*p* + 1) (*p* − 4) = (*p* + (−2)) (*p*
+ l) (*p* + (−4))

Comparing (*p* − 2) (*p* + 1) (*p* − 4) with (*x*
+ *a*) (*x* + *b*) (*x* + *c*) we have *x* = *p*
; *a* = −2; *b* = 1 ; *c* = −4.

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

= *p*^{3} + (−2 + 1 + (−4)) *p*^{2} +
((−2) (1) + (1) (−4) + (−4) (−2))*p* + (−2) (1) (−4)

= *p*^{3} + (−5) *p*^{2} + (−2 + (−4)
+ 8) *p* + 8

= *p*^{3} − 5*p*^{2} + 2*p* + 8

** **

**5. Find the volume of the cube whose
side is ( x **

**Solution:**

Given side of the cube = (*x* + 1) cm

Volume of the cube = (side)^{3} cubic units = (*x*
+ 1)^{3} cm^{3}

We have (*a + b*)^{3} = (*a*^{3} + 3*a*^{2}*b*
+ 3*ab*^{2} + *b*^{3}) cm^{3}

(*x* + 1)^{3} = (*x*^{3} + 3*x*^{2}
(1) + 3*x* (1)^{2} + l^{3}) cm^{3}

Volume = (*x*^{3} + 3*x*^{2} + 3*x*
+ 1) cm^{3}

** **

**6. Find the volume of the cuboid whose
dimensions are ( x **

**Solution:**

Given the dimensions of the cuboid as = (*x* + 2), (*x* – 1) and (*x*
– 3)

∴ Volume of the cuboid = (*l* × *b* × *h*) units^{3}

= (*x* + 2) (*x* − 1) (*x* − 3) units^{3}

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

∴ (*x* + 2) (*x* − l) (*x* − 3) = *x*^{3}
+ (2 − 1 − 3)*x*^{2} + (2 (−1) + (−1) (−3) + (−3) (2)) *x*
+(2)(−l) (−3)

= *x*^{3} – 2*x*^{2} + (−2 + 3 − 6)*x*
+ 6

Volume = *x*^{3} – 2*x*^{2} − 5*x*
+ 6 units^{3}

** **

**Objective
Type Questions**

**7. If x^{2}–y^{2}
= 16 and (x+y) = 8 then (x–y) is ___________**

(A) 8

(B) 3

(C) 2

(D) 1

**[Answer: (C) 2]**

**Solution: **

*x*^{2}− *y*^{2} = 16

(*x − y*) (*x* − *y*) = 16

8 (*x* − *y*) = 16

(*x* − *y*) = 16 / 8 = 2

**8. **

(A) *a*^{2}–*ab*+*b*^{2}

(B) *a*^{2}+*ab*+*b*^{2}

(C) *a*^{2}+2*ab*+*b*^{2}

(D) *a*^{2}–2*ab*+*b*^{2}

**[Answer: (B) a^{2} + ab + b^{2 }]**

**Solution:**

[(*a* − *b*)^{2} (*a*^{3} – *b*^{3})
] / [(*a*^{2} – *b*^{2})] = [ (*a* + *b*) (*a* – *b*)
*a*^{2} + *ab* + *b*^{2 })] / [ (*a* + *b*)
(*a* – *b*)]

= *a*^{2} + *ab*
+ *b*^{2}

**9. ( p+q)(p^{2}–pq+q^{2}) is equal to _____________**

(A) *p*^{3}+*q*^{3}

(B) (*p*+*q*)^{3}

(C) *p*^{3}–*q*^{3}

(D) (*p*–*q*)^{3}

**[Answer: (A) p^{3} + q^{3 }]**

**Solution:**

*a*^{3} + *b*^{3 }=
(*a* + *b*)(*a*^{2} − *ab* + *b*^{2})

**10. ( a–b)=3 and ab=5 then a^{3}–b^{3}
= ____________**

(A) 15

(B) 18

(C) 62

(D) 72

**[Answer: (D) 72]**^{}

**Solution:**

(*a* − *b*) = 3

(*a* − *b*)^{2} = 3^{2}

*a*^{2} + *b*^{2} − 2*ab* = 9

*a*^{2} + *b*^{2} – 2(5)
= 9

*a*^{2} + *b*^{2} = 9 + 10

*a*^{2} + *b*^{2 }= 19

*a*^{3} + *b*^{3} = (*a* – *b*) (*a*^{2}
+ *ab* + *b*^{2}) = 3 (19 + 5)

= 3(24) = 72

**11. a^{3}+b^{3} = (a+b)^{3} – __________**

(A) 3*a*(*a*+*b*)

(B) 3*ab*(*a*–*b*)

(C) –3*ab*(*a*+*b*)

(D) 3*ab*(*a*+*b*)

**[Answer:** **3 ab (a + b)]**

**Solution:**

(*a + b*)^{3} = *a*^{3 }+ *b*^{3}
+ 3*a*^{2}*b* + 3*ab*^{2}

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

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

**Answer**

**Exercise 3.3 **

**1. (i) 9m ^{2} +
30 m + 25 (ii) 25 p^{2}
− 10p +1 (iii ) 4n^{2}
+ 4n − 3 (iv) ( 2p + 5q )( 2p − 5q)**

**2. (i) m^{3} + 9m^{2} + 27 m + 27 (ii) 8a^{3} + 60 a^{2} +150 a +125 (iii ) 27 p^{3} + 108 p^{2}q +144 pq^{2} + 64q^{3} (iv) 14,0608 (v) 1124864 **

**3. (i) 125 − 75 x +15x^{2} − x^{3}
(ii) 8x^{3} − 48x^{2} y + 96 xy^{2} − 64y^{3} (iii ) a^{3}b^{3} − 3a^{2}b^{2}c + 3abc^{3} − c^{3}**

**(iv) 110, 592 ( v ) 912673 x^{3}y^{3} **

**4. p^{3} − 5p^{2} + 2p + 8 **

**5. x^{3} + 3x^{2}
+ 3 x +1 **

**6. x^{3} − 2x^{2}
− 5 x + 6**

**7. (C) 2 **

**8. (B) a ^{2} +
ab + b^{2} **

**9. (A) p^{3} + q^{3} **

**10. (D) 72 **

**11. (D) 3 ab(a+b)**

Tags : Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

8th Maths : Chapter 3 : Algebra : Exercise 3.3 (Cubic Identities) | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.