If a-b=3 and b-c=5 then find the value of a²+b²+c²-ab-bc-ca.
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a-b= 3⇒ a= 3+b ··· (1)
b-c =5 ⇒ b= 5+c ···(2)
From (1) and (2) we get,
a = 3+5+c = 8+c •••(3)
Now,
a + b + c - ab - bc - ca =
(8+c)²+(5+c)²+c² -(8+c)(5+c) - (5+c)(c) -(c)(8+c)
=(64 + 16c +c²) + (25 + 10c + c²) + c² -(40+8c+5c+c²) - (5c+c²) - (8c+c²)
=(64+25+ 26c + 3c²) - (40+26c+3c²)
=(89 - 40)+ (26c -26c) + (3c² - 3c²)
=89 - 40 =49
So the anwer is 49.
b-c =5 ⇒ b= 5+c ···(2)
From (1) and (2) we get,
a = 3+5+c = 8+c •••(3)
Now,
a + b + c - ab - bc - ca =
(8+c)²+(5+c)²+c² -(8+c)(5+c) - (5+c)(c) -(c)(8+c)
=(64 + 16c +c²) + (25 + 10c + c²) + c² -(40+8c+5c+c²) - (5c+c²) - (8c+c²)
=(64+25+ 26c + 3c²) - (40+26c+3c²)
=(89 - 40)+ (26c -26c) + (3c² - 3c²)
=89 - 40 =49
So the anwer is 49.
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