Question

If a²+b²+c²=83 and a+b+c=15, then find the value of ab+bc+ca.

Answer 1

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)

(15)^2=83+2(ab+bc+ca)

225–83=2(ab+bc+ca)

142/2=71=ab+bc+ca

a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

=(15)(83–71)

=15×12

=180 , Answer

Answer 2

Given:

• a² + b² + c² = 83

• a + b + c = 15

To find out:

Find the value of ab + bc + ca ?

Solution:

We know that,

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

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

Putting the values in the above formula, we get

⇒ ( 15 )² = 83 + 2 ( ab + bc + ca )

⇒ 225 - 83 = 2 ( ab + bc + ca )

⇒ 2 ( ab + bc + ca ) = 142

⇒ ab + bc + ca = 142/2

⇒ ab + bc + ca = 71

Some useful identities:

• ( a - b )³ = a³ - b³ - 3ab ( a - b )

• a³ + b³ = ( a + b ) ( a² - ab + b² )

• a³ - b³ = ( a - b ) ( a² + ab + b² )

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