Question

if a+b+c=2, 1/a+1/b+1/c=0, ac= 4/5 and a^3+ b^3+ c^3=28 find the value of a^2+ b^ 2+c^2
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Answer 1

Step-by-step explanation:

(a+b+c)=10;a

2

+b

2

+c

2

=38 & a

3

+b

3

+c

3

=160

(a+b+c)

2

=a

2

+b

2

+c

2

+2(ab+bc+ac)⇒

2

100−38

=ab+bc+ca

⇒ab+bc+ca=31

As a

3

+b

3

+c

3

−3abc=(a+b+c)[a

2

+b

2

+c

2

−ab−bc−ca]

160−3abc=10[38−31]

⇒

3

160−70

=abc⇒abc=30

Answer 2

Answer:

$\sf \: a+b+c=9 and a2+b2+c2=35</p><p></p><p>$

Using formula,

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

$\sf92=35+2(ab+bc+ca)$

$\sf2(ab+bc+ca)=81−35=46$

$\sf(ab+bc+ca)=23$

$\sf \: using \: formula, </p><p>(a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c</p><p>$ \sfa3+b3+c3−3abc=(35−23)×9=9×12=108[/tex]

$\sf \: Answer  108$