**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 (a ^{2} – bc) + b (b^{2 }– c) + c (c^{2} – ab) = 0 **

L.H.S = a (a^{2} – bc) + b (b^{2 }– c) + c (c^{2} – ab)

= a^{3} – abc + b^{3} – abc + c^{3} – abc

= a^{3} + b^{3} + c^{3} – 3abc

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

a^{3} + b^{3} + c^{3} = 3abc

Hence we have

= 3abc – 3abc

= 0 = R.H.S

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

L.H.S = a (b^{2} + c^{2}) + b (c^{2} + a^{2}) + c (a^{2} + b^{2})

= ab^{2} + ac^{2}+ bc^{2} + ba^{2} + ca^{2} +cb^{2}

= ab^{2} + ba^{2} + b^{2}c + bc^{2} + ac^{2} + a^{2}c

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

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

= (ab)^{2} + (bc)^{2} + (ca)^{2} +2ab.bc + ^{2}bc. ca + ^{2}ca.ab

= a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} + 2ab^{2}c + ^{2}bc^{2}a + ^{2}ca^{2}b

= a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} + 2abc + (0)

= a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2} = R.H.S

**(v) a ^{2} – bc = b^{2} – ca = c^{2} – ab = – (ab + bc + ca) **

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

b^{2} – ca = b.b – ca = b(-c-a)-ca = -bc-ab-ca = -(ab+bc+ca) ———-(2)

c^{2} – ab = c.c – ab = c(-a-b) – ab = -ac-bc-ab = -(ab+bc+ca) ———-(3)

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

a^{2} – bc = b^{2} – ca = c^{2} – ab = – (ab + bc + ca)

**(vi) 2a ^{2} + bc = (a – b) (a –c) **

L.H.S = 2a^{2 }+ bc

= a^{2} + a^{2} + bc

= a^{2} + a x a + bc

= a^{2} + a (- b – c) + bc

= a^{2} – 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) a ^{2} + b^{2} + c^{2} = -2(ab + bc + ca) **

We have a + b + c = 0

Squaring we get

(a + b + c)^{2} = 0

a^{2} + b^{2} + c^{2} + 2ab + 2bc +2ca = 0

a^{2} + b^{2} + c^{2} = – 2ab – 2bc – 2ca

a^{2} + b^{2} + c^{2} = – 2(ab + bc + ca)

Hence the proof

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

**Prove the following: **

**(i) ****𝐚 ^{𝟐}**

**/**

_{𝐛𝐜}**+**

**𝐛**

^{𝟐}**/**

_{𝐜𝐚}**+**

**𝐜**

^{𝟐}**/**

_{𝐚𝐛}**= 3**

L.H.S = a^{2}/_{bc} + b^{2}/_{ca} + c^{2}/_{ab}

= (a^{2}.a+b^{2}.b+c^{2}.c)/_{abc}

= (a^{3}+b^{3}+c^{3})/_{abc} ………. (1)

We have a + b + c = 0, a + b = –c

Cubing we get

(a + b)^{3} = (–c)^{3}

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

a^{3} + b^{3} – 3ab = –c^{3}

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

**/**

_{𝟐𝐚}

^{𝟐}

_{+}

_{𝐛𝐜}**+**

**𝐛**

^{𝟐}**/**

_{𝟐𝐛}

^{𝟐}

_{𝐜𝐚}**+**

**𝐜**

^{𝟐}**/**

_{𝟐𝐜}

^{𝟐}

_{𝐚𝐛}**= 1, provided the denominators do not become 0.**

L.H.S = 𝐚^{𝟐}/_{𝟐𝐚}^{𝟐}_{+}_{𝐛𝐜} + 𝐛^{𝟐}/_{𝟐𝐛}^{𝟐}_{𝐜𝐚} + 𝐜^{𝟐}/_{𝟐𝐜}^{𝟐}_{𝐚𝐛}

= 𝐚^{𝟐}/_{(a-b)(a-c) }+ 𝐛^{𝟐}/_{(b-c)(b-a)} + 𝐜^{𝟐}/_{(c-a)(c-b)}

= 𝐚^{𝟐}/_{(a-b)(a-c) }– 𝐛^{𝟐}/_{(a-b)(a-c} + 𝐜^{𝟐}/_{(a-c)(b-c) }

= [a^{2}(b−c)– b^{2}(a−c) + c^{2}(a−b)] /_{(a−b)(b−c)(a−c) }

= [a^{2}b− a^{2}c − b^{2}a + b^{2}c + c^{2}a− c^{2}b] /_{(ab− b}^{2}_{− ac +bc) (a−c)}

= [a^{2}b− a^{2}c − b^{2}a + b^{2}c + c^{2}a− c^{2}b]/[a^{2}b − ab^{2}− a^{2}c + abc − abc + b^{2}c + ac^{2}]

= 1 = R.H.S

**If a + b + c = 0, prove that b**^{2}– 4ac is a square.

We have a + b + c = 0

b = – (a + c)

Squaring on both sides

b^{2} = [- (a + c)]^{2}

= (a + c)^{2}

b^{2} = a^{2} + c^{2} + 2ac

Subtracting 4ac on both sides

b^{2} – 4ac = a^{2} + c^{2} + 2ac – 4ac

= a^{2} – 2ac + c^{2}

b^{2} – 4ac = (a – c)^{2}

We find that b^{2} – 4ac is the square of (a – c)

**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) = s ^{2} **

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

= s^{2} – as + s^{2} – bs + s^{2} – cs

= 3s^{2} – as – bs – cs

= 3s^{2} – s (a + b + c)

= 3s^{2} – s (2s) (a + b + c = 2s)

= 3s^{2} – 2s^{2}

= s^{2} = R.H.S.

**(ii) s ^{2} (s – a)^{2} + s (s – b)^{2} + s (s – c)^{2} = a^{2} + b^{2} + c^{2} **

L.H.S. = s^{2} (s – a)^{2} + s (s – b)^{2} + s (s – c)^{2}

= s^{2} + s^{2} + a^{2} – ^{2}sa + s^{2} + b^{2} + s^{2} + c^{2} – 2as – 2bs – 2cs

= 4s^{2} + a^{2} + b^{2} + c^{2} – 2s – 2bs – 2cs

= 4s^{2} + a^{2} + b^{2} + c^{2} – 2s (a + b + c)

= 4s^{2} + a^{2} + b^{2} + c^{2} – 2as (2s) (a + b + c = 2s)

= 4s^{2} + a^{2} + b^{2} + c^{2} – 4s^{2}

= a^{2} + b^{2} + c^{2}

= R.H.S.

**(iii) (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s ^{2} = ab + bc + ca **

L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s^{2}

= s^{2} – as – bs + ab + s^{2} – bs – cs + bc + s^{2} – cs – as + ac + s^{2}

= 4s^{2} – 2 as – 2bs – 2cs + ab + bc +ca

= 4s^{2} – 2s (a + b + c) + ab + bc +ca

= 4s^{2} – 2s (2s) + ab + bc +ca

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

= ab + bc +ca

= R.H.S.

**(iv) a ^{2} – b^{2} – c^{2} + 2bc = 4 (s – b) ( s – c) **

L.H.S = a^{2} – b^{2} – c^{2} + 2bc

= a^{2} – (b^{2} + c^{2} – 2bc)

= a^{2} – (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.

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

**Prove that **^{𝐚}**/ _{(}**

_{𝐬}

_{−}

_{𝐚}

_{)}**+**

^{𝐛}**/**

_{(}

_{𝐬}

_{−}

_{𝐛}

_{)}**+**

^{𝐜}**/**

_{(}

_{𝐬}

_{−}

_{𝐜}

_{)}**+ 2 =**

^{𝐚𝐛𝐜}**/**

_{(}

_{𝐬}

_{−}

_{𝐜}

_{)( }

_{𝐬}

_{−}

_{𝐛}

_{) (}

_{𝐬}

_{−}

_{𝐜}

_{)}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(s^{2}−bs−cs+bc) +b(s^{2}−as−cs+ac) +c(s^{2}−as−bs+ab) +2(s^{2}−as−bs+ab)(s−c)] /_{(s−a)(s−b)(S−c) }

= [as^{2}−abs−acs+abc+bs^{2}−abc−bcs+abc) +cs^{2}−acs−bcs+abc+2(s^{3}−as^{2}−bs^{2}+abs −cs^{2}+acs+bcs−abc] /_{(s−a)(s−b)(s−c) }

= [s^{2}(a+b+c) − 2abc −2acs − 2bcs − 3abc + 2s^{3}−2as^{2}− 2bs^{2} + 2abs−2cs^{2}+ 2acs+2bcs−2abc] /_{(s−a)(s−b)(s−c)}

= [s^{2}(2s) + abc+2s^{3}−2s^{2} (a+b+c)]/_{(s−a)(s−b)(S−c)}

=[2s^{3}+ abc+2s^{3}−2s^{2}(2s)]/_{(s−a) s−b (S−c)}

=[4s^{3}+ abc−4s^{3}]/_{(s−a)(s−b)(S−c)}

= ^{abc}/_{(s−a)(s−b)(S−c)}

= R.H.S.

**If a + b + c = 0, prove that a**^{2}– bc = b^{2 }– ca = c^{2}– ab =**(****𝐚**^{𝟐}**+****𝐛**^{𝟐}**+****𝐜**^{𝟐}**)/**_{𝟐}

We have a + b + c = 0

Squaring will get

(a + b + c)^{2} = 0

a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 0

a^{2} + b^{2} + c^{2} + 2b (a + c) + 2ca = 0

a^{2} + b^{2} + c^{2} + 2b (-b) + 2ca = 0

a^{2} + b^{2} + c^{2} = 2b^{2} – 2ca (hint: a + c = -b)

a^{2} + b^{2} + c^{2} = 2 (b^{2} – ca)

a^{2} + b^{2} + c^{2} = b^{2} – ca ……… (1)

IIIly (a + b + c)^{2} = 0

a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 0

a^{2} + b^{2} + c^{2} + 2ab + 2c (b + a) = 0

a^{2} + b^{2} + c^{2} + 2ab + 2c (-c) = 0 (a + c = -c)

a^{2} + b^{2} + c^{2} = 2 (c^{2} – ab)

[a^{2} + b^{2} + c^{2}]/_{2} = (c^{2} – ab)……………(2)

Also a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 0

a^{2} + b^{2} + c^{2} + 2a (b + c) + 2bc = 0

a^{2} + b^{2} + c^{2} + 2a (-a) + 2bc = 0

a^{2} + b^{2} + c^{2} = 2 (a^{2} – bc) ……..(3)

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

a^{2} – bc = b^{2} – ca = c^{2} – ab = [a^{2} + b^{2} + c^{2}]/_{2}

**If 2(a**^{2}+ b^{2}) = (a + b)^{2}, prove that a = b.

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

2a^{2} + 2b^{2} – a^{2} – b^{2} – 2ab = 0

a^{2} + b^{2} – 2ab = 0

(a – b)^{2} = 0

a – b =0

a = b

**If x**^{2}– 3x + 1 = 0, prove that x^{2}+^{𝟏}**/****𝐱**^{𝟐}**= 7.**

Given x^{2} – 3x + 1 = 0

x^{2} + 1 = 3x

x + ^{1}/_{x} = 3 (dividing both sides by x)

Squaring both sides we get

(x + ^{1}/_{x} )^{2} = 3^{2} = x^{2} + ^{1}/_{x} ^{2} + 2x.^{ 1}/_{x} = 9

x^{2} + ^{1}/_{x} ^{2} + 2 = 9

x^{2} + ^{1}/_{x} ^{2} = 9 – 2 = 7

## One thought on “Multiplication of Polynomials Exercise 3.1.5-Class 9”

Comments are closed.