Question

If a+b+c=15 and ab+bc+ca=71. Find value of a2+b2+c2

Answer 1

Answer:

taking a2+b2+c2 as x

$({a + b + c})^{2} = {a}^{2} + {b}^{2} + {c}^{2} \\ + 2(ab + bc + ca) \\ 225 = {x} + 142 \\ x = 83$

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Answer 2

hay!!

question

$if \: a + b + c = 15 \: and \: ab + bc + ca = 71 \: then \: find \: the \: value \: of \: a {}^{2} b {}^{2} c {}^{2}$

given

a + b + c = 15

ab + bc + ca = 71

to find

$a {}^{2}+ b {}^{2}+ c {}^{2}$

answer

083

formulas used

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

${\sf{\red{\underline{\Large{Explanation}}}}}$

formula

=>(a+b+c)²=a²+b²+c²+2(ab + bc + ac)

putting the values

=>(a+b+c)²=a²+b²+c²+2(ab + bc + ac)

=>(15)²=a²+b²+c²+2(71)

=>225 =a²+b²+c²+142

=>225-142=a²+b²+c²

=>083. answer.

hope it's helps you