Question

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If a + b + c = 9 and ab + bc + ca = 23, then
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Answer 1

Answer:

a²+b²+c² = 35

Step-by-step explanation:

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

a+b+c= 9 (given)

ab+ bc + ca = 23 (given)

⇒ a³+b³+c³- 3abc = 9 (a²+b²+c²- 23)            ... (1)

(a+b+c)²= a²+b²+c²+ 2ab+ 2bc+ 2ca

92= a²+b²+c²+ 2(ab+bc+ca)

81= a²+b²+c²+ 2(23)

81 = a²+b²+c²+ 46

a²+b²+c²= 81-46

a²+b²+c² = 35

Answer 2

Answer:

We are given with a+b+c= 9 and ab+bc+ca =23

=> 2(ab+bc+ac)=46

=>2ab+2bc+2ac=46

Now,

a+b+c=9

Squaring on both sides we get

=> $(a+b+c)^2=9^2$

=> $a^2+b^2+c^2+ab+2bc+2ac=81$

=> $a^2+b^2+c^2+46=81$ ( since 2ab+2bc+2ca=46)

Therefore  $a^2+b^2+c^2=35$.

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