Question : If a²+ b² + c²= ab + bc + ac, then the value of a+c/b is
1. 6
2. 2
3. 1
4. - 1
Answers
Answered by
0
Step-by-step explanation:
a
3
+ b
3
+ c
3
- 3abc
= (a
2
+ b
2
+ c
2
- ab - bc - ca)(a + b + c)
= a
2
+ b
2
+ c
2
= ab + bc + ca (given)
⇒ (a
2
+ b
2
+ c
2
- ab - bc - ca) = 0
∴ a
3
+ b
3
+ c
3
- 3abc
(a + b + c) x 0 = 0
∴ a
3
+ b
3
+ c
3
= 3abc
Answered by
0
Answer:
2
Step-by-step explanation:
We know that sum of perfect squares is greater than or equal to zero
So,
(a−b)²+(b−c)²+(c−a)²>=0
So , expanding this we get,
a²+b²+c²>=ab+bc+ac
But in the question the reverse inequality is given
a²+b²+c² < =ab+bc+ac
Hence it follows that ,
a²+b²+c²=ab+bc+ac
So ,
(a−b)²+(b−c)²+(c−a)²=0
Hence,
a=b=c
So a+bc=2
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