Question

if a + b + c equal to 15 and a b + BC + CA = 271 then find the value of a square + b square + c square ​

Answer 1

Answer:

-317

Step-by-step explanation:

Identity!

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

Thus, the answer is $15^2-271*2=225-542=-317$.

Answer 2

Application of Algebraic Identity

Answer: if a + b + c = 15 and ab + bc + ca = 271 then a²+b²+c² = -317.

Explanation:

Given that a + b + c = 15 and ab + bc + ca = 271

Need to find the value of a²+b²+c²

Value of a²+b²+c² can be obtained by using following algebraic identity.

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

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

On substituting a + b + c = 15 and ab + bc + ca = 217 in above expression we get

(15)² = a²+b²+c² + 2(271)

=> 225 = a²+b²+c² + 542

=> 225 - 542 = a²+b²+c²

=> a²+b²+c² = -317

Hence if a + b + c = 15 and ab + bc + ca = 271 then a²+b²+c² = -317.

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