Question

ab+bc+ca=71 and a+b+c=15 find a^2 + b^2 + c^2

Answer 1

GIVEN:

ab+bc+ca=71

a+b+c=15

Then,

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

we know,

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

Then,

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

=(15)^2-2(71)

=225-142

=083

HOPE THIS WILL HELP YOU

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