Question 1: Write the following in the expanded form:
(i) (a + 2b + c)2
(ii) (2a โ 3b โ c)2
(iii) (โ3x+y+z)2
(iv) (m+2nโ5p)2
(v) (2+xโ2y)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) (2xโy+z)2
(xii) (โ2x+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 โ 3b โ c)2
= [(2a) + (โ3b) + (โc)]2
= (2a)2 + (โ3b)2 + (โc)2 + 2(2a)(โ3b) + 2(โ3b)(โc) + 2(2a)(โc)
= 4a2 + 9b2 + c2 โ 12ab + 6bc โ 4ca
(iii) (โ3x+y+z)2
= [(โ3x)2 + y2 + z2 + 2(โ3x)y + 2yz + 2(โ3x)z]
= 9x2 + y2 + z2 โ 6xy + 2yz โ 6xz
(iv) (m+2nโ5p)2
= m2 + (2n)2 + (โ5p)2 + 2m ร 2n + (2ร2nรโ5p) + 2m ร (โ5p)
= m2 + 4n2 + 25p2 + 4mn โ 20np โ 10pm
(v) (2+xโ2y)2
= 22 + x2 + (โ2y) 2 + 2(2)(x) + 2(x)(โ2y) + 2(2)(โ2y)
= 4 + x2 + 4y2 + 4 x โ 4xy โ 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
=(\frac{x}{y})^2+(\frac{y}{z})^2+(\frac{z}{x})^2+2\frac{x}{y}ร\frac{y}{z}+2\frac{y}{z}ร\frac{z}{x}+2\frac{z}{x}ร\frac{x}{y}\\=(\frac{x^2}{y^2})+(\frac{y^2}{z^2})+(\frac{z^2}{x^2})+2\frac{x}{z}+2\frac{y}{x}+2\frac{z}{y} (ix) (a/bc + b/ac + c/ab)2
=(\frac{a}{bc})^2+(\frac{b}{ca})^2+(\frac{c}{ab})^2+2(\frac{a}{bc})(\frac{b}{ca})+2(\frac{b}{ca})(\frac{c}{ab})+2(\frac{a}{bc})(\frac{c}{ab})\\=(\frac{a^2}{b^2c^2})+(\frac{b^2}{c^2a^2})+(\frac{c^2}{a^2b^2})+\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^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) (2xโy+z)2
= (2x)2 + (โy)2 + (z)2 + 2(2x)(โy) + 2(โy)(z) + 2(2x)(z)
= 4x2 + y2 + z2 โ 4xy โ 2yz + 4xz
(xii) (โ2x+3y+2z)2
= (โ2x)2 + (3y)2 + ( 2z)2 + 2(โ2x)(3y) + 2(3y)(2z) + 2(โ2x)(2z)
= 4x2 + 9y2 + 4z2 โ12xy + 12yz โ8xz
Question 2: Simplify
(i) (a + b + c)2 + (a โ b + c)2
(ii) (a + b + c)2 โ (a โ b + c)2
(iii) (a + b + c)2 + (a โ b + c)2 + (a + b โ c)2
(iv) (2x + p โ c)2 โ (2x โ p + c)2
(v) (x2 + y2 โ z2)2 โ (x2 โ y2 + z2)2
Solution:
(i) (a + b + c)2 + (a โ b + c)2
= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (โb)2 + c2 โ2abโ2bc+2ca)
= 2a2 + 2 b2 + 2c2 + 4ca
(ii) (a + b + c)2 โ (a โ b + c)2
= (a2 + b2 + c2 + 2ab+2bc+2ca) โ (a2 + (โb)2 + c2 โ2abโ2bc+2ca)
= a2 + b2 + c2 + 2ab + 2bc + 2ca โ a2 โ b2 โ c2 + 2ab + 2bc โ 2ca
= 4ab + 4bc
(iii) (a + b + c)2 + (a โ b + c)2 + (a + b โ c)
= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c)2 โ 2ab โ 2cb + 2ca) + (a2 + b2 + c2 + 2ab โ 2bc โ 2ca)
= 3a2 + 3b2 + 3c2 + 2ab โ 2bc + 2ca
(iv) (2x + p โ c)2 โ (2x โ p + c)2
= [4x2 + p2 + c2 + 4xp โ 2pc โ 4xc] โ [4x2 + p2 + c2 โ 4xpโ 2pc + 4xc]
= 4x2 + p2 + c2 + 4xp โ 2pc โ 4cx โ 4x2 โ p2 โ c2 + 4xp + 2pcโ 4cx
= 8xp โ 8xc
= 8(xp โ xc)
(v) (x2 + y2 โ z2)2 โ (x2 โ y2 + z2)2
= (x2 + y2 + (โz)2)2 โ (x2 โ y2 + z2)2
= [x4 + y4 + z4 + 2x2y2 โ 2y2z2 โ 2x2z2 โ [x4 + y4 + z4 โ 2x2y2 โ 2y2z2 + 2x2z2]
= 4x2y2 โ 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
