Verified answer

Answer:

To find the value of ab + bc + ca we can expand and simplify the equation a² + b² + c² = (a + b + c)².

Expanding the right side of the equation:

(a + b + c)² = (a + b + c)(a + b + c)

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

= a² + ab + ac + ab + b² + bc + ac + bc + c²

= a² + b² + c² + 2ab + 2ac + 2bc

Since we know that a² + b² + c² = (a + b + c)² we can substitute this in:

a² + b² + c² = (a + b + c)²

=> a² + b² + c² = a² + b² + c² + 2ab + 2ac + 2bc

We can then simplify the equation:

0 = 2ab + 2ac + 2bc

=> -2ab - 2ac - 2bc = 0

=> -2(ab + ac + bc) = 0

=> ab + ac + bc = 0

Therefore the value of ab + ac + bc is 0. So the answer is option 1) 0.

Answer:

Solution

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Consider, a

2

+b

2

+c

2

–ab–bc–ca=0

Multiply both sides with 2, we get

2(a

2

+b

2

+c

2

–ab–bc–ca)=0

⇒ 2a

2

+2b

2

+2c

2

–2ab–2bc–2ca=0

⇒ (a

2

–2ab+b

2

)+(b

2

–2bc+c

2

)+(c

2

–2ca+a

2

)=0

⇒ (a–b)

2

+(b–c)

2

+(c–a)

2

=0

Since the sum of square is zero then each term should be zero

⇒ (a–b)

2

=0,(b–c)

2

=0,(c–a)

2

=0

⇒ (a–b)=0,(b–c)=0,(c–a)=0

⇒ a=b,b=c,c=a

∴ a=b=c.