Question

If a²+b²+c²= 2(3a-4b-6c)-61, then the value lof a-b-c is​

Answer 1

Answer:

here is your answer mate!!!!

Answer 2

The value of a - b - c = 13

Given :

a² + b² + c² = 2(3a - 4b - 6c) - 61

To find :

The value of a - b - c

Solution :

Step 1 of 3 :

Write down the given equation

The given equation is

a² + b² + c² = 2(3a - 4b - 6c) - 61

Step 2 of 3 :

Find the value of a, b, c

$\displaystyle \sf{ {a}^{2} + {b}^{2} + {c}^{2} = 2(3a - 4b - 6c) - 61}$

$\displaystyle \sf{ \implies {a}^{2} + {b}^{2} + {c}^{2} = 6a - 8b - 12c - 61}$

$\displaystyle \sf{ \implies {a}^{2} - 6a + {b}^{2} + 8b+ {c}^{2} + 12c + 61= 0}$

$\displaystyle \sf{ \implies {a}^{2} - 6a + 9+ {b}^{2} + 8b+ 16+ {c}^{2} + 12c + 36= 0}$

$\displaystyle \sf{ \implies ({a}^{2} - 2.a.3 + {3}^{2} ) + ( {b}^{2} + 2.b.4+ {4}^{2} ) + ( {c}^{2} + 2.c.6 + {6}^{2}) = 0}$

$\displaystyle \sf{ \implies {(a - 3)}^{2} + {(b + 4)}^{2} + {(c + 6)}^{2} = 0}$

We know that if sum of squares of three Real Numbers are zero then they are separately zero

Thus we get

a - 3 = 0 , b + 4 = 0 , c + 6 = 0

Consequently we get

a = 3 , b = - 4 , c = - 6

Step 3 of 3 :

Find the value of a - b - c

a - b - c

= 3 + 4 + 6

= 13

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