# Multiplication of Polynomials Exercise 3.1.5-Class 9

1. If a + b + c = 0, prove the following:

(i) (b + c) (b – c) + a (a + 2b) = 0

Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have

L.H.S = (b + c) (b – c) + a (a + 2b)

= (-a) (b – c) + a (a + b + b)

= -ab + ac + a (-c + b)

= -ab + ab – ac + ac

= 0= R.H.S

(ii) a (a2 – bc) + b (b2 – c) + c (c2 – ab) = 0

L.H.S = a (a2 – bc) + b (b2 – c) + c (c2 – ab)

= a3 – abc + b3 – abc + c3 – abc

= a3 + b3 + c3 – 3abc

We know that if a + b + c = 0 then

a3 + b3 + c3 = 3abc

Hence we have

= 3abc – 3abc

= 0 = R.H.S

(iii) a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = –3abc

L.H.S = a (b2 + c2) + b (c2 + a2) + c (a2 + b2)

= ab2 + ac2+ bc2 + ba2 + ca2 +cb2

= ab2 + ba2 + b2c + bc2 + ac2 + a2c

= ab (a + b) + bc (b + c) + ac (a + c)

= ab (–c) + bc (–a) + ac (–b)

= –abc – abc – abc

= –3abc = R.H.S

[a + b + c =0, a + b = -c, b + c = -a, c + a = -b]

(iv) (ab + bc + ca)2 = a2b2 + b2c2 + c2a2

L.H.S = (ab + bc + ca)2

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

= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b

= a2b2 + b2c2 + c2a2 + 2abc + (0)

= a2b2 + b2c2 + c2a2 = R.H.S

(v) a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)

a.a – bc = a(-b-c)-bc = -ab-ac-ba = -(ab+bc+ca)——(1)

b2 – ca = b.b – ca = b(-c-a)-ca = -bc-ab-ca = -(ab+bc+ca) ———-(2)

c2 – ab = c.c – ab = c(-a-b) – ab = -ac-bc-ab = -(ab+bc+ca) ———-(3)

From equation (1) (2) and (3)

a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)

(vi) 2a2 + bc = (a – b) (a –c)

L.H.S = 2a2 + bc

= a2 + a2 + bc

= a2 + a x a + bc

= a2 + a (- b – c) + bc

= a2 – ab – ac + bc

= a (a – b) – c (a – b)

= (a – b) (a –c) = R.H.S

(vii) (a + b) (a – b) + ca – cb = 0

We have a + b + c = 0

a + b = c

L.H.S = (a + b) (a – b) + ac – cb

= –c (a – b) + ac – cd

= – ca + bc + ac – cb

= 0 = R.H.S

(viii) a2 + b2 + c2 = -2(ab + bc + ca)

We have a + b + c = 0

Squaring we get

(a + b + c)2 = 0

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

a2 + b2 + c2 = – 2ab – 2bc – 2ca

a2 + b2 + c2 = – 2(ab + bc + ca)

Hence the proof

1. Suppose a, b, c are non-zero real numbers such that a + b + c = 0,

Prove the following:

(i) 𝐚𝟐/𝐛𝐜 + 𝐛𝟐/𝐜𝐚 + 𝐜𝟐/𝐚𝐛 = 3

L.H.S = a2/bc + b2/ca + c2/ab

= (a2.a+b2.b+c2.c)/abc

= (a3+b3+c3)/abc ………. (1)

We have a + b + c = 0, a + b = –c

Cubing we get

(a + b)3 = (–c)3

a3 + b3 + 3ab + (a + b) = –c3

a3 + b3 – 3ab = –c3

a3 + b3 + c3 = 3abc ………. (2)

Substituting (2) in (1)

L.H.S = 3abc/abc = 3

(ii) ( +𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/𝐛 ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )

Whenever b + c ≠ 0, c + a ≠ 0, a + b ≠ 0

We have a + b + c =0

a + b = –c

b + c = –a

c + a = –b

L.H.S = (𝐚+𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/ ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )

= ( −c/c + −a/a + −a/b ) ( b/−b + c/−c + a/−a )

= (-1-1-1) (-1-1-1)

= (-3) (-3)

= 9 = R.H.S

(iii) 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛 = 1, provided the denominators do not become 0.

L.H.S = 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛

= 𝐚𝟐/(a-b)(a-c) + 𝐛𝟐/(b-c)(b-a) + 𝐜𝟐/(c-a)(c-b)

= 𝐚𝟐/(a-b)(a-c) – 𝐛𝟐/(a-b)(a-c + 𝐜𝟐/(a-c)(b-c)

= [a2(b−c)– b2(a−c) + c2(a−b)] /(a−b)(b−c)(a−c)

= [a2b− a2c − b2a + b2c + c2a− c2b] /(ab− b2− ac +bc) (a−c)

= [a2b− a2c − b2a + b2c + c2a− c2b]/[a2b − ab2− a2c + abc − abc + b2c + ac2]

= 1 = R.H.S

1. If a + b + c = 0, prove that b2 – 4ac is a square.

We have a + b + c = 0

b = – (a + c)

Squaring on both sides

b2 = [- (a + c)]2

= (a + c)2

b2 = a2 + c2 + 2ac

Subtracting 4ac on both sides

b2 – 4ac = a2 + c2 + 2ac – 4ac

= a2 – 2ac + c2

b2 – 4ac = (a – c)2

We find that b2 – 4ac is the square of (a – c)

1. If a, b, c are real numbers such that a + b + c = 2s, prove the following:

(i) s (s – a) + s (s – b) + s (s – c) = s2

L.H.S. = s (s – a) + s (s – b) + s (s – c)

= s2 – as + s2 – bs + s2 – cs

= 3s2 – as – bs – cs

= 3s2 – s (a + b + c)

= 3s2 – s (2s) (a + b + c = 2s)

= 3s2 – 2s2

= s2 = R.H.S.

(ii) s2 (s – a)2 + s (s – b)2 + s (s – c)2 = a2 + b2 + c2

L.H.S. = s2 (s – a)2 + s (s – b)2 + s (s – c)2

= s2 + s2 + a22sa + s2 + b2 + s2 + c2 – 2as – 2bs – 2cs

= 4s2 + a2 + b2 + c2 – 2s – 2bs – 2cs

= 4s2 + a2 + b2 + c2 – 2s (a + b + c)

= 4s2 + a2 + b2 + c2 – 2as (2s) (a + b + c = 2s)

= 4s2 + a2 + b2 + c2 – 4s2

= a2 + b2 + c2

= R.H.S.

(iii) (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2 = ab + bc + ca

L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2

= s2 – as – bs + ab + s2 – bs – cs + bc + s2 – cs – as + ac + s2

= 4s2 – 2 as – 2bs – 2cs + ab + bc +ca

= 4s2 – 2s (a + b + c) + ab + bc +ca

= 4s2 – 2s (2s) + ab + bc +ca

= 4s2 – 4s2 + ab + bc +ca (a + b + c = 2s)

= ab + bc +ca

= R.H.S.

(iv) a2 – b2 – c2 + 2bc = 4 (s – b) ( s – c)

L.H.S = a2 – b2 – c2 + 2bc

= a2 – (b2 + c2 – 2bc)

= a2 – (b – c)2

= (a + b – c) [a – (b – c)]

= (a + b – c) (a + b – c)

= (2s – c – c) (2s – b – b)

= (2s – 2c) (2s – 2b)

= 2(s – c) (2) (s – b)

= 4 (s – c) (s – b)

= R.H.S.

1. If a, b, c are real numbers, a + b + c =2s and s – a ≠ 0, s – b ≠ 0, s – c ≠ 0,

Prove that 𝐚/(𝐬𝐚) + 𝐛/(𝐬𝐛) + 𝐜/(𝐬𝐜) + 2 = 𝐚𝐛𝐜/(𝐬𝐜)( 𝐬𝐛) (𝐬𝐜)

L.H.S = a/(s−a) + b/(s−b) + c/(s−c) + 2

=a(s−b)( s−c) + b(s−a)(s−c) + c(s−a)( s−b) + 2(s−a)(s−b)(s−c)/ (s−a)(s−b)(s−c)

= [a(s-b)(s-c)+b(s-a)(s-c) +c(s-a)(s-b)+2(s-a)(s-b)(s-c)] /(s−a )(s−b)(S−c)

=[a(s2−bs−cs+bc) +b(s2−as−cs+ac) +c(s2−as−bs+ab) +2(s2−as−bs+ab)(s−c)] /(s−a)(s−b)(S−c)

= [as2−abs−acs+abc+bs2−abc−bcs+abc) +cs2−acs−bcs+abc+2(s3−as2−bs2+abs −cs2+acs+bcs−abc] /(s−a)(s−b)(s−c)

= [s2(a+b+c) − 2abc −2acs − 2bcs − 3abc + 2s3−2as2− 2bs2 + 2abs−2cs2+ 2acs+2bcs−2abc] /(s−a)(s−b)(s−c)

=  [s2(2s) + abc+2s3−2s2 (a+b+c)]/(s−a)(s−b)(S−c)

=[2s3+ abc+2s3−2s2(2s)]/(s−a) s−b (S−c)

=[4s3+ abc−4s3]/(s−a)(s−b)(S−c)

= abc/(s−a)(s−b)(S−c)

= R.H.S.

1. If a + b + c = 0, prove that a2 – bc = b2 – ca = c2 – ab = (𝐚𝟐 + 𝐛𝟐+𝐜𝟐)/𝟐

We have a + b + c = 0

Squaring will get

(a + b + c)2 = 0

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

a2 + b2 + c2 + 2b (a + c) + 2ca = 0

a2 + b2 + c2 + 2b (-b) + 2ca = 0

a2 + b2 + c2 = 2b2 – 2ca                 (hint: a + c = -b)

a2 + b2 + c2 = 2 (b2 – ca)

a2 + b2 + c2 = b2 – ca ……… (1)

IIIly (a + b + c)2 = 0

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

a2 + b2 + c2 + 2ab + 2c (b + a) = 0

a2 + b2 + c2 + 2ab + 2c (-c) = 0 (a + c = -c)

a2 + b2 + c2 = 2 (c2 – ab)

[a2 + b2 + c2]/2 = (c2 – ab)……………(2)

Also a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

a2 + b2 + c2 + 2a (b + c) + 2bc = 0

a2 + b2 + c2 + 2a (-a) + 2bc = 0

a2 + b2 + c2 = 2 (a2 – bc) ……..(3)

From (1), (2) and (3) we get,

a2 – bc = b2 – ca = c2 – ab = [a2 + b2 + c2]/2

1. If 2(a2 + b2) = (a + b)2, prove that a = b.

2a2 + 2b2 = a2 + b2 + 2ab

2a2 + 2b2 – a2 – b2 – 2ab = 0

a2 + b2 – 2ab = 0

(a – b)2 = 0

a – b =0

a = b

1. If x2 – 3x + 1 = 0, prove that x2 + 𝟏/𝐱𝟐 = 7.

Given x2 – 3x + 1 = 0

x2 + 1 = 3x

x + 1/x = 3 (dividing both sides by x)

Squaring both sides we get

(x + 1/x )2 = 32 = x2 + 1/x 2 + 2x. 1/x = 9

x2 + 1/x 2 + 2 = 9

x2 + 1/x 2 = 9 – 2 = 7