Question

25. If a2 +b2 + c2 = 64 and ab + bc + ca = 18, find a +b+c.
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Answer 1

Given:-

=>a²+b²+c²=64

=>ab+bc+ca=18

To find:-

=>Value of a+b+c

Solution:-

=>(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

=>(a+b+c)²=64+2(18)

=>(a+b+c)²=100

=>a+b+c=10

Thus,value of a+b+c is 10.

Answer 2

Answer :

$\large\boxed{\sf{\star\:\:a\:+\:b\:+\:c\:=\:10\:\:\star}}$

Explanation :

Correct Question :–

If a² + b² + c² = 64 and ab + bc + ca = 18, find a + b + c .

$\dag$ Given :–

• a² + b² + c² = 64
• ab + bc +ca = 18

$\dag$ To Find :–

a + b + c = ?

$\dag$ Formula Applied :–

• (a + b + c)² = a² + b² + c² + 2(ab + bc +ca)

$\dag$ Solution :–

 We are given , a² + b² + c² = 64 and ab + bc + ca = 18 .

 We also know that (a + b + c)² = a² + b² + c² + 2(ab + bc +ca).

 Putting the above values in this formula, We get :

⇒ (a + b + c)² = (a² + b² + c²) + 2(ab + bc +ca)

⇒ (a + b + c)² = (64) + [2 × (18)]

⇒ (a + b + c)² = 64 +36

⇒ (a + b + c)² = 100

⇒ a + b + c = √(100)

⇒ a + b + c = 10

∴ Sum of a , b and c will be 10 .

$\bigstar$ Important Formulae :–

• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)
• a³ + b³ + c³ - 3ab = (a + b + c)(a² + b² +c² - ab - bc - ca)

[Note : If a+ b + c = 0 then the formula will be a³ + b³ +c³ = 3abc .]