MCQ
If $a^2+b^2+c^2-a b-b c-c a=0$, then:
- A$a+b+c$
- B$b+c=a$
- C$c+a=b$
- ✓$a=b=c$
✓
Answer
Correct option: D.
$a=b=c$
$a^2+b^2+c^2-a b-b c-c a=0$
Multiplying by $2$ on both the sides, we have
$2\left(a^2+b^2+c^2-a b-b c-c a\right)=0$
$2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a=0$
$a^2+a^2+b^2+b^2+c^2+c^2-2 a b-2 b c-2 c a=0$
$\left(a^2+b^2-2 a b\right)+\left(b^2+c^2-2 b c\right)+\left(a^2+c^2-2 a c\right)=0$
$(a-b)^2+(b-c)^2+(a-c)^2=0$
$(a-b)^2=0,(b-c)^2=0,(a-c)^2=0$
$(a-b)=0,(b-c)=0,(a-c)=0$
$a=b, b=c, a=c$
or we can say $a = b = c$
Hence, correct option is $(d).$
Multiplying by $2$ on both the sides, we have
$2\left(a^2+b^2+c^2-a b-b c-c a\right)=0$
$2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a=0$
$a^2+a^2+b^2+b^2+c^2+c^2-2 a b-2 b c-2 c a=0$
$\left(a^2+b^2-2 a b\right)+\left(b^2+c^2-2 b c\right)+\left(a^2+c^2-2 a c\right)=0$
$(a-b)^2+(b-c)^2+(a-c)^2=0$
$(a-b)^2=0,(b-c)^2=0,(a-c)^2=0$
$(a-b)=0,(b-c)=0,(a-c)=0$
$a=b, b=c, a=c$
or we can say $a = b = c$
Hence, correct option is $(d).$
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