Question

if a-b=3 and b-c=5 then the value of a square+b square +c square - ab-bc-ca is

Answer 1

a-b = 3……..(1)

b-c= 5 ………….(2)

On adding eq 1& 2

a-c= 8 …………….(3)

From eq 1
a-b= 3
a= 3+b……………(4)

From eq 2
b-c = 5

b = 5+c…………….(5)


From eq 3

a-c = 8

c= a-8……………(6)

a²+b²+c²-ab-bc-ca (given)

= a²-ab +b²-bc+c²-ca

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

=a(3) + b (5) + c (-8)

[ From eq 1, 2,3]

=3a+5b -8c

3a + 5(5+c) -8(a-8)

[ From eq 5,6]

=3a + 5 (5+a-8) -8a+64. ( c=a-8)

=3a + 25 +5a -40-8a+64

=3a+5a-8a+25-40+64

=8a -8a-15+64

a²+b²+c²-ab-bc-ca = 49

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Hope this will help you.....

Answer 2

To Find  a²+b²+c²-ab-bc-ca

Given :
a-b=3 
b-c=5 

To Find  a²+b²+c²-ab-bc-ca

Solution :

a -b =3---- equation (1)

b-c =5 --- equation (2)
  
 Adding equation 1 and 2:

a - b = 3
b - c = 5
+   +    +
________

a -c =8 -------equation (3)

⇒ c = a -8 

a -b =3---- equation (1)

⇒a=b+3

b-c =5 --- equation (2)
 ⇒b=c+5

Now let us find

 a²+b²+c²-ab-bc-ca= a²-ab +b² -bc +c² -ca
 
=a(a-b)+b(b-c)+c(c-a)
=a(3) + b(5) +c(-8)

=3a +5b -8c

=3a +5[c+5]-8[a-8]

=3a+5[a-8+5] -8a+64

=3a +5[a-3]-8a+64

=3a+5a -15 +64 -8a

=8a-8a +49
=49

∴a²+b²+c²-ab-bc-ca =49