If a-b=3 and b-c=5 then find a^2-b^2-c^2-ab-bc-ca
Answers
Answered by
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
==================================================================
Hope this will help you....
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
==================================================================
Hope this will help you....
Answered by
0
Dear student I think, the question should have been
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
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
Similar questions