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# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.2

• Last Updated : 28 Dec, 2020

### Question 1: Write the following in the expanded form:

(i) (a + 2b + c)2

(ii) (2a − 3b − c)2

(iii) (−3x+y+z)2

(iv) (m+2n−5p)2

(v) (2+x−2y)2

(vi) (a2 +b2 +c2)2

(vii) (ab+bc+ca)2

(viii) (x/y+y/z+z/x)2

(ix) (a/bc + b/ac + c/ab)2

(x) (x+2y+4z)2

(xi) (2x−y+z)2

(xii) (−2x+3y+2z)2

Solution:

By following the identity,

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz

(i) (a + 2b + c)2

= a2 + (2b)2 + c2 + 2a(2b) + 2ac + 2(2b)c

= a2 + 4b2 + c2 + 4ab + 2ac + 4bc

(ii) (2a − 3b − c)2

= [(2a) + (−3b) + (−c)]2

= (2a)2 + (−3b)2 + (−c)2 + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c)

= 4a2 + 9b2 + c2 − 12ab + 6bc − 4ca

(iii) (−3x+y+z)2

= [(−3x)2 + y2 + z2 + 2(−3x)y + 2yz + 2(−3x)z]

= 9x2 + y2 + z2 − 6xy + 2yz − 6xz

(iv) (m+2n−5p)2

= m2 + (2n)2 + (−5p)2 + 2m × 2n + (2×2n×−5p) + 2m × (−5p)

= m2 + 4n2 + 25p2 + 4mn − 20np − 10pm

(v) (2+x−2y)2

= 22 + x2 + (−2y) 2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y)

= 4 + x2 + 4y2 + 4 x − 4xy − 8y

(vi) (a2 +b2 +c2)2

= (a2)2 + (b2)2 + (c2)2 + 2a2 b2 + 2b2c2 + 2a2c2

= a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2

(vii) (ab+bc+ca)2

= (ab)2 + (bc)2 + (ca)2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)

= a2b2 + b2c2 + c2a2 + 2(ac)b2 + 2(ab)(c)2 + 2(bc)(a)2

(viii) (x/y+y/z+z/x)2

(ix) (a/bc + b/ac + c/ab)2

(x) (x+2y+4z)2

= x2 + (2y)2 + (4z)2 + (2x)(2y) + 2(2y)(4z) + 2x(4z)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(xi) (2x−y+z)2

= (2x)2 + (−y)2 + (z)2 + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z)

= 4x2 + y2 + z2 − 4xy − 2yz + 4xz

(xii) (−2x+3y+2z)2

= (−2x)2 + (3y)2 + ( 2z)2 + 2(−2x)(3y) + 2(3y)(2z) + 2(−2x)(2z)

= 4x2 + 9y2 + 4z2 −12xy + 12yz −8xz

### Question 2: Simplify

(i) (a + b + c)2 + (a − b + c)2

(ii) (a + b + c)2 − (a − b + c)2

(iii) (a + b + c)2 + (a – b + c)2 + (a + b − c)2

(iv) (2x + p − c)2 − (2x − p + c)2

(v) (x2 + y2 − z2)2 − (x2 − y2 + z2)2

Solution:

(i) (a + b + c)2 + (a − b + c)2

= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (−b)2 + c2 −2ab−2bc+2ca)

= 2a2 + 2 b2 + 2c2 + 4ca

(ii) (a + b + c)2 − (a − b + c)2

= (a2 + b2 + c2 + 2ab+2bc+2ca) − (a2 + (−b)2 + c2 −2ab−2bc+2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca

= 4ab + 4bc

(iii) (a + b + c)2 + (a – b + c)2 + (a + b − c)

= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c)2 − 2ab − 2cb + 2ca) + (a2 + b2 + c2 + 2ab − 2bc – 2ca)

= 3a2 + 3b2 + 3c2 + 2ab − 2bc + 2ca

(iv) (2x + p − c)2 − (2x − p + c)2

= [4x2 + p2 + c2 + 4xp − 2pc − 4xc] − [4x2 + p2 + c2 − 4xp− 2pc + 4xc]

= 4x2 + p2 + c2 + 4xp − 2pc − 4cx − 4x2 − p2 − c2 + 4xp + 2pc− 4cx

= 8xp − 8xc

= 8(xp − xc)

(v) (x2 + y2 − z2)2 − (x2 − y2 + z2)2

= (x2 + y2 + (−z)2)2 − (x2 − y2 + z2)2

= [x4 + y4 + z4 + 2x2y2 – 2y2z2 – 2x2z2 − [x4 + y4 + z4 − 2x2y2 − 2y2z2 + 2x2z2]

= 4x2y2 – 4z2x2

### Question 3: If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.

Solution:

Given,

a + b + c = 0 and a2 + b2 + c2 = 16

Choose a + b + c = 0

Squaring both sides,

(a + b + c)2 = 0

a2 + b2 + c2 + 2(ab + bc + ca) = 0

16 + 2(ab + bc + c) = 0

2(ab + bc + ca) = -16

ab + bc + ca = -16/2 = -8

or

ab + bc + ca = -8

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